Undeclared first use in this function error in c

This article by Scaler Topics explains errors & their types in C, covering the explanation & examples for each type of error in C Programming Language.

Types of Errors in C

Overview

An error in the C language is an issue that arises in a program, making the program not work in the way it was supposed to work or may stop it from compiling as well.
If an error appears in a program, the program can do one of the following three things: the code will not compile, the program will stop working during execution, or the program will generate garbage values or incorrect output. There are five different types of errors in C Programming like Syntax Error, Run Time Error, Logical Error, Semantic Error, and Linker Error.

Scope

  • This article explains errors and their types in C Programming Language.
  • This article covers the explanation and examples for each type of error in C Programming Language (syntax error, run time error, logical error, semantic error, linker error).

Introduction

Let us say you want to create a program that prints today’s date. But instead of writing printf in the code, you wrote print. Because of this, our program will generate an error as the compiler would not understand what the word print means. Hence, today’s date will not print. This is what we call an error. An error is a fault or problem in a program that leads to abnormal behavior of the program. In other words, an error is a situation in which the program does something which it was not supposed to do. This includes producing incorrect or unexpected output, stopping a program that was running, or hindering the code’s compilation. Therefore it is important to remove all errors from our code, this is known as debugging.

How to Read an Error in C?

In order to resolve an error, we must figure out how and why did the error occur. Whenever we encounter an error in our code, the compiler stops the code compilation if it is a syntax error or it either stops the program’s execution or generates a garbage value if it is a run time error.

Syntax errors are easy to figure out because the compiler highlights the line of code that caused the error. Generally, we can find the error’s root cause on the highlighted line or above the highlighted line.

For example:

#include <stdio.h>
int main() {
    int var = 10
    return 0;
}

Output:

error: expected ',' or ';' before 'return'
      4 |  return 0;

As we can see, the compiler shows an error on line 4 of the code. So, in order to figure out the problem, we will go through line 4 and a few lines above it. Once we do that, we can quickly determine that we are missing a semi-colon (;) in line 4. The compiler also suggested the same thing.

Other than the syntax errors, run time errors are often encountered while coding. These errors are the ones that occur while the code is being executed.

Let us now see an example of a run time error:

#include<stdio.h>

void main() {
    
    int var;
    var = 20 / 0;
    
    printf("%d", var);
}

Output:

warning: division by zero [-Wdiv-by-zero]
    6 |     var = 20 / 0;

As we can see, the compiler generated a warning at line 6 because we are dividing a number by zero.

Sometimes, the compiler does not throw a run time error. Instead, it returns a garbage value. In situations like these, we have to figure out why did we get an incorrect output by comparing the output with the expected output. In other cases, the compiler does not display any error at all. The program execution just ends abruptly in cases like these.

Let us take another example to understand this kind of run time error:

#include <stdio.h>
#include <stdlib.h>

int main() {
    
	int arr[1]; 
	arr[0] = 10; 

	int val = arr[10000]; 
	printf("%d", val); 
    return 0;
}

Output:


In the above code, we are trying to access the 10000th element but the size of array is only 1 therefore there is no space allocated to the 10000th element, this is known as segmentation fault.

Types of Errors in C

There are five different types of errors in C.

  1. Syntax Error
  2. Run Time Error
  3. Logical Error
  4. Semantic Error
  5. Linker Error

1. Syntax Error

Syntax errors occur when a programmer makes mistakes in typing the code’s syntax correctly or makes typos. In other words, syntax errors occur when a programmer does not follow the set of rules defined for the syntax of C language.

Syntax errors are sometimes also called compilation errors because they are always detected by the compiler. Generally, these errors can be easily identified and rectified by programmers.

The most commonly occurring syntax errors in C language are:

  • Missing semi-colon (;)
  • Missing parenthesis ({})
  • Assigning value to a variable without declaring it

Let us take an example to understand syntax errors:

#include <stdio.h>

void main() {
    var = 5;    // we did not declare the data type of variable
     
    printf("The variable is: %d", var);
}

Output:

error: 'var' undeclared (first use in this function)

If the user assigns any value to a variable without defining the data type of the variable, the compiler throws a syntax error.

Let’s see another example:

#include <stdio.h>

void main() {
    
    for (int i = 0;) {  // incorrect syntax of the for loop 
        printf("Scaler Academy");
    }
}

Output:

error: expected expression before ')' token

A for loop needs 3 arguments to run. Since we entered only one argument, the compiler threw a syntax error.

2. Runtime Error

Errors that occur during the execution (or running) of a program are called RunTime Errors. These errors occur after the program has been compiled successfully. When a program is running, and it is not able to perform any particular operation, it means that we have encountered a run time error. For example, while a certain program is running, if it encounters the square root of -1 in the code, the program will not be able to generate an output because calculating the square root of -1 is not possible. Hence, the program will produce an error.

Runtime errors can be a little tricky to identify because the compiler can not detect these errors. They can only be identified once the program is running. Some of the most common run time errors are: number not divisible by zero, array index out of bounds, string index out of bounds, etc.

Runtime errors can occur because of various reasons. Some of the reasons are:

  1. Mistakes in the Code: Let us say during the execution of a while loop, the programmer forgets to enter a break statement. This will lead the program to run infinite times, hence resulting in a run time error.
  2. Memory Leaks: If a programmer creates an array in the heap but forgets to delete the array’s data, the program might start leaking memory, resulting in a run time error.
  3. Mathematically Incorrect Operations: Dividing a number by zero, or calculating the square root of -1 will also result in a run time error.
  4. Undefined Variables: If a programmer forgets to define a variable in the code, the program will generate a run time error.

Example 1:

// A program that calculates the square root of integers
#include <stdio.h>
#include <math.h>

int main() {
    for (int i = 4; i >= -2; i--)     {
        printf("%f", sqrt(i));
        printf("n");
    }      
    return 0;
}

Output:

2.000000
1.732051
1.414214
1.000000
0.000000
-1.#IND00
-1.#IND00

**In some compilers, you may also see this output: **

2.000000
1.732051
1.414214
1.000000
0.000000
-nan
-nan

In the above example, we used a for loop to calculate the square root of six integers. But because we also tried calculating the square root of two negative numbers, the program generated two errors (the IND written above stands for «Indeterminate»). These errors are the run time errors.
-nan is similar to IND.

Example 2:

#include<stdio.h>
 
void main() {
    int var = 2147483649;

    printf("%d", var);
}

Output:


This is an integer overflow error. The maximum value an integer can hold in C is 2147483647. Since in the above example, we assigned 2147483649 to the variable var, the variable overflows, and we get -2147483647 as the output (because of the circular property).

3. Logical Error

Sometimes, we do not get the output we expected after the compilation and execution of a program. Even though the code seems error free, the output generated is different from the expected one. These types of errors are called Logical Errors. Logical errors are those errors in which we think that our code is correct, the code compiles without any error and gives no error while it is running, but the output we get is different from the output we expected.

In 1999, NASA lost a spacecraft due to a logical error. This happened because of some miscalculations between the English and the American Units. The software was coded to work for one system but was used with the other.

For Example:

#include <stdio.h>

void main() {
    float a = 10;
    float b = 5;
    
    if (b = 0) {  // we wrote = instead of ==
        printf("Division by zero is not possible");
    } else {
        printf("The output is: %f", a/b);
    }
}

Output:


INF signifies a division by zero error. In the above example, at line 8, we wanted to check whether the variable b was equal to zero. But instead of using the equal to comparison operator (==), we use the assignment operator (=). Because of this, the if statement became false and the value of b became 0. Finally, the else clause got executed.

4. Semantic Error

Errors that occur because the compiler is unable to understand the written code are called Semantic Errors. A semantic error will be generated if the code makes no sense to the compiler, even though it is syntactically correct. It is like using the wrong word in the wrong place in the English language. For example, adding a string to an integer will generate a semantic error.

Semantic errors are different from syntax errors, as syntax errors signify that the structure of a program is incorrect without considering its meaning. On the other hand, semantic errors signify the incorrect implementation of a program by considering the meaning of the program.

The most commonly occurring semantic errors are: use of un-initialized variables, type compatibility, and array index out of bounds.

Example 1:

#include <stdio.h>

void main() {
    int a, b, c;
    
    a * b = c;
    // This will generate a semantic error
}

Output:

error: lvalue required as left operand of assignment

When we have an expression on the left-hand side of an assignment operator (=), the program generates a semantic error. Even though the code is syntactically correct, the compiler does not understand the code.

Example 2:

#include <stdio.h>

void main() {
    int arr[5] = {5, 10, 15, 20, 25};
    
    int arraySize = sizeof(arr)/sizeof(arr[0]);
    
    for (int i = 0; i <= arraySize; i++)
    {
        printf("%d n", arr[i]);
    }
}

Output:


In the above example, we printed six elements while the array arr only had five. Because we tried to access the sixth element of the array, we got a semantic error and hence, the program generated a garbage value.

5. Linker Error

Linker is a program that takes the object files generated by the compiler and combines them into a single executable file. Linker errors are the errors encountered when the executable file of the code can not be generated even though the code gets compiled successfully. This error is generated when a different object file is unable to link with the main object file. We can run into a linked error if we have imported an incorrect header file in the code, we have a wrong function declaration, etc.

For Example:

#include <stdio.h>
 
void Main() { 
    int var = 10;
    printf("%d", var);
}

Output:

undefined reference to `main'

In the above code, as we wrote Main() instead of main(), the program generated a linker error. This happens because every file in the C language must have a main() function. As in the above program, we did not have a main() function, the program was unable to run the code, and we got an error. This is one of the most common type of linker error.

Conclusion

  • There are 5 different types of errors in C programming language: Syntax error, Runtime error, Logical error, Semantic error, and Linker error.
  • Syntax errors, linker errors, and semantic errors can be identified by the compiler during compilation. Logical errors and run time errors are encountered after the program is compiled and executed.
  • Syntax errors, linker errors, and semantic errors are relatively easy to identify and rectify compared to the logical and run time errors. This is so because the compiler generates these 3 (syntax, linker, semantic) errors during compilation itself, while the other 2 errors are generated during or after the execution.

Improve Article

Save Article

  • Read
  • Discuss
  • Improve Article

    Save Article

    Variables: A variable is the name given to a memory location. It is the basic unit of storage in a program.

    • The value stored in a variable can be changed during program execution.
    • A variable is only a name given to a memory location, all the operations done on the variable effects that memory location.
    • All the variables must be declared before use.

    How to declare variables?
    We can declare variables in common languages (like C, C++, Java etc) as follows:
    Variables in Java

    where:
    datatype: Type of data that can be stored in this variable.
    variable_name: Name given to the variable.
    value: It is the initial value stored in the variable.
    

    How to avoid errors while creating variables?

    1. The identifier is undeclared: In any programming language, all variables have to be declared before they are used. If you try to use the name of a such that hasn’t been declared yet, an “undeclared identifier” compile-error will occur.

      Example:

      #include <stdio.h>

      int main()

      {

          printf("%d", x);

          return 0;

      }

      Compile Errors:

      prog.c: In function 'main':
      prog.c:5:18: error: 'x' undeclared (first use in this function)
           printf("%d", x);
                        ^
      prog.c:5:18: note: each undeclared identifier is reported
           only once for each function it appears in
      
    2. No initial value is given to the variable: This error commonly occurs when the variable is declared, but not initialized. It means that the variable is created but no value is given to it. Hence it will take the default value then. But in C language, this might lead to error as this variable can have a garbage value (or 0) as its default value. In other languages, 0 will be its default value.

      Example:

      #include <stdio.h>

      int main()

      {

          int x;

          printf("%d", x);

          return 0;

      }

      Output:

      0
      
    3. Using variable out of its Scope: Scope of a variable is the part of the program where the variable is accessible. Like C/C++, in Java, all identifiers are lexically (or statically) scoped, i.e.scope of a variable can be determined at compile time and independent of the function call stack.

      Example:

      #include <stdio.h>

      int main()

      {

          {

              int x = 5;

          }

          printf("%d", x);

          return 0;

      }

      Compile Errors:

      prog.c: In function 'main':
      prog.c:5:18: error: 'x' undeclared (first use in this function)
           printf("%d", x);
                        ^
      prog.c:5:18: note: each undeclared identifier is reported
           only once for each function it appears in
      

      How to Correct the above code: Declare the variable x before using it in the outer scope. Or you can use the already defined variable x in its own scope

      Example:

      #include <stdio.h>

      int main()

      {

          {

              int x = 5;

              printf("%d", x);

          }

          return 0;

      }

      Output:

      5
      
    4. Creating a variable with an incorrect type of value: This arises due to the fact that values are implicitly or explicitly converted into another type. Sometimes this can lead to Warnings or errors.

      Example:

      #include <stdio.h>

      int main()

      {

          char* x;

          int i = x;

          printf("%d", x);

          return 0;

      }

      Warning:

      prog.c: In function 'main':
      prog.c:7:13: warning: initialization makes integer
       from pointer without a cast [-Wint-conversion]
           int i = x;
                   ^
      

    Очень часто начинающие программисты впадают в суеверный ужас, когда видят, что компилятор нашел в тексте программы ошибку, но не понимают, в чем она заключается.

    А если помножить этот факт на незнание английского языка («чего там ему не нравится?..») и слабое владение синтаксисом C++ («хм, а может, тут нужна точка с запятой…»), то проблема принимает масштаб катастрофы.

    Тот факт, что компилятор в силу своих ограниченных возможностей изо всех сил старается объяснить, что конкретно неверно, не спасает ситуацию. Как быть, если гуглить неохота, а спросить не у кого?

    В этом посте на правах копипаста с последующим переводом, дополнениями и исправлениями приведу описание наиболее распространенных сообщений об ошибках и предупреждений компилятора. Неприятность кроется в том факте, что разные компиляторы ругаются на одинаковые ошибки по-разному, а некоторые даже не замечают то, что другие принимают за ошибку. Все зависит от совести разработчиков компилятора, даты его выпуска, и др.

    В качестве компилятора возьмем g++, который, в частности, может использоваться в среде Code::Blocks. Версия gcc (куда входит g++) для ОС Windows зовется MinGW. По ходу я буду давать аналоги ошибок из лексикона русскоязычной Microsoft Visual C++.

    Итак, частые ошибки:

    undeclared identifier

    1) Пример

    doy.cpp: In function 'int main()':
    doy.cpp:25: 'DayOfYear' undeclared (first use this function)
    doy.cpp:25: (Each undeclared identifier is reported only once for each function it appears in.)
    doy.cpp:25: parse error before ';' token

    2) Смысл
    Использован идентификатор DayOfYear, но компилятор не нашел его объявления. Он не знает, что такое DayOfYear.

    3) Когда бывает

    • Вы забыли включить какой-то заголовочный файл (#include...)
    • Вы где-то ошиблись в написании идентификатора (при объявлении или использовании)
    • Вы вообще забыли, что эту переменную надо объявить

    Попытавшись скомпилировать это в Microsoft Visual C++, вы увидите:

    error C2065: DayOfYear: необъявленный идентификатор

    cout undeclared

    1) Пример

    xyz.cpp: In function 'int main()':
    xyz.cpp:6: 'cout' undeclared (first use this function)
    xyz.cpp:6: (Each undeclared identifier is reported only once for each function it appears in.)

    2) Смысл
    Суперклассика. Без комментариев.

    3) Когда бывает

    • Вы забыли включить <iostream>
    • Вы забыли написать using namespace std;

    jump to case label

    1) Пример

    switch.cpp: In function 'int main()':
    switch.cpp:14: jump to case label
    switch.cpp:11: crosses initialization of 'int y'

    2) Смысл
    Смысл туманен

    3) Когда бывает
    Вы попытались объявить и инициализировать переменную (объект, указатель и т.п.) в метке case оператора выбора switch. Правилами C++ это запрещено.

    В Microsoft Visual C++ эта ошибка зовется

    error C2360: пропуск инициализации 'y' из-за метки 'case'

    Выход: заключите операторы этого case’а в фигурные скобки {}.

    multi-line string / unterminated string

    1) Пример
    Программка

    #include <iostream>

    using namespace std;

    int main()
    {
    cout << "Bob is my buddy;
    cout << "and so is Mary" << endl;
    }

    вызовет бурную реакцию компилятора:

    string.cpp:7:12: warning: multi-line string literals are deprecated
    string.cpp: In function 'int main()':
    string.cpp:7: 'so' undeclared (first use this function)
    string.cpp:7: (Each undeclared identifier is reported only once for each function it appears in.)
    string.cpp:7: parse error before 'Mary'
    string.cpp:8:28: warning: multi-line string literals are deprecated
    string.cpp:8:28: missing terminating " character
    string.cpp:7:12: possible start of unterminated string literal

    2) Смысл
    Компилятор думает, что мы хотим создать строковую константу с содержащимся в ней переносом строки, что-то типа

    "Hello
    world!"

    что не поддерживается языком. Также делается предположение о том, что мы, возможно, забыли поставить кавычки в конце первой строки. Собственно, так оно и есть.

    3) Когда бывает
    Когда не соблюдается правильное количество и положение кавычек в строковых литералах. Надо быть внимательнее.

    Microsoft Visual C++ со свойственной ему детской непосредственностью, отметит, что нельзя делать переносы в строках и возмутится, где точка с запятой:

    error C2001: newline в константе
    error C2146: синтаксическая ошибка: отсутствие ";" перед идентификатором "cout"

    comparison between signed and unsigned integer expressions

    1) Пример

    xyz.cpp: In function 'int main()':
    xyz.cpp:54: warning: comparison between signed and unsigned integer expressions

    2) Смысл
    Это — предупреждение компилятора, которое говорит о том, что мы пытаемся сравнить (==, и т.д.) целочисленное выражение (может принимать положительные, отрицательные значения и 0) и беззнаковое целочисленное выражение (может быть только положительным, либо 0).

    3) Когда бывает
    Собственно, тогда и бывает. Напомню, что тип int по умолчанию знаковый, а некоторые функции (например, vector::size()) возвращают unsigned int.
    К примеру, следующий на первый взгляд безобидный код вызовет описываемое предупреждение:

    for (int i = 0; i < grades.size(); i++)
    {
    // ...
    }

    Следует помнить, что в памяти знаковые и беззнаковые типы имеют разные внутренние представления, поэтому надо быть чертовски осторожными с указателями.

    В Microsoft Visual C++ предупреждение выглядит так:

    warning C4018: <: несоответствие типов со знаком и без знака

    suggest parentheses around assignment used as truth value

    1) Пример

    xyz.cpp: In function `int main()':
    xyz.cpp:54: warning: suggest parentheses around assignment used as truth value

    2) Смысл
    Тоже классика. Компилятор предполагает (и в 99% случаев прав), что вы по ошибке включили в скобки в качестве условия для if/while/for вместо условного выражения выражение присваивания.

    3) Когда бывает
    Чаще всего — в if‘ах, когда вместо "==" используется "="

    if (length = maxLength)

    вместо

    if (length == maxLength)

    Заминка в том, что это не ошибка, т.к. в скомпилированной программе (если мы проигнорируем предупреждение) выражение присваивания (которое возвращает значение правого аргумента) во всех случаях, кроме тех, когда оно вернет 0, будет преобразовано к true.

    Ссылки для дальнейшего изучения
    Ошибки построения Microsoft Visual C++
    GCC Compiler error messages
    GCC Warnings

    P.S. Следует отметить, что кроме ошибок стадии компиляции встречаются (гораздо реже) ошибки препроцессора (например, если не найден заголовочный файл <iostram>), ошибки стадии компоновки (можно избежать, если научиться пользоваться средой программирования) и — самый гнусный тип ошибок! — ошибки стадии выполнения. Т.н. runtime error. С ними может справиться только голова программиста, вооруженная отладчиком.

    ASORGOO

    0 / 0 / 0

    Регистрация: 11.01.2020

    Сообщений: 47

    1

    29.05.2021, 17:05. Показов 1732. Ответов 4

    Метки си (Все метки)


    Здравствуй форум. Столкнулся с такой проблемой:
    Имеется программа для проверки атрибутов файла и вывод их на экран.
    Но программа не работает — при попытке компиляции выскакивает
    [Error] ‘FA_RDONLY’ undeclared (first use in this function) на 27 строке
    [Error] ‘FA_HIDDEN’ undeclared (first use in this function) на 29 строке
    [Error] ‘FA_SYSTEM’ undeclared (first use in this function) на 31 строке
    [Error] ‘FA_LABEL’ undeclared (first use in this function) на 33 строке
    [Error] ‘FA_DIREC’ undeclared (first use in this function) на 35 строке
    [Error] ‘FA_ARCH’ undeclared (first use in this function) на 37 строке
    Библиотеки проверил, все целые, на всякий случай скачал и заменил с интернета. В чём проблема и что можно сделать с этим?
    Компилировал в DEV-C++ на Windows 7 Максимальной

    C
    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    13
    14
    15
    16
    17
    18
    19
    20
    21
    22
    23
    24
    25
    26
    27
    28
    29
    30
    31
    32
    33
    34
    35
    36
    37
    38
    39
    40
    41
    42
    43
    44
    45
    46
    
    #include<errno.h>
    #include<stdio.h>
    #include<dos.h>
    #include<io.h>
     
    int get_file_attrib(char *filename);
     
    int main(void)
    {
       char filename[128];
       int attrib;
       printf("Введите имя файла:");
       scanf("%s",filename);
       attrib = get_file_attrib(filename);
       if(attrib == -1)
          switch(errno)
          {
             case ENOENT:printf("Маршрут или имя файла не найденыn");
                         break;
             case EACCES:printf("Отказ доступаn");
                         break;
             default:    printf("Ошибка номер %dn",errno);
                         break;
          }
       else
       {
          if(attrib & FA_RDONLY)
             printf("%s имеет атрибут только на чтениеn",filename);
          if(attrib & FA_HIDDEN)
             printf("%s - cкрытый файлn",filename);
          if(attrib & FA_SYSTEM)
             printf("%s - системный файл n",filename);
          if(attrib & FA_LABEL)
             printf("%s - метка томаn",filename);
          if(attrib & FA_DIREC)
             printf("%s - каталогn",filename);
          if(attrib & FA_ARCH)
             printf("%s - архивный файлn",filename);
       }
       return 0;
    }
    /*  возвращает атрибуты файла */
    int get_file_attrib(char *filename)
    {
       return(_chmod(filename,0));
    }

    __________________
    Помощь в написании контрольных, курсовых и дипломных работ, диссертаций здесь



    0



    2083 / 918 / 428

    Регистрация: 17.11.2018

    Сообщений: 2,405

    29.05.2021, 17:48

    2



    0



    0 / 0 / 0

    Регистрация: 11.01.2020

    Сообщений: 47

    29.05.2021, 18:16

     [ТС]

    3

    Так то да, это работающий код, но он на C++. А я писал на Си. Хоть вроде сильных отличий между языками и кодами нет, anyway не работает



    0



    analogov net

    2083 / 918 / 428

    Регистрация: 17.11.2018

    Сообщений: 2,405

    29.05.2021, 18:30

    4

    Лучший ответ Сообщение было отмечено ASORGOO как решение

    Решение

    Цитата
    Сообщение от ASORGOO
    Посмотреть сообщение

    А я писал на Си.

    ASORGOO, вот на Си:

    C
    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    13
    14
    15
    16
    17
    18
    19
    20
    21
    22
    23
    24
    25
    26
    27
    28
    29
    30
    31
    32
    33
    34
    35
    36
    37
    38
    39
    40
    
    int main( void )
    {
        SetConsoleCP( 1251 );
        SetConsoleOutputCP( 1251 );
     
        char filename[128];
        DWORD attrib;
        printf( "Введите имя файла:" );
        scanf( "%s", filename );
     //   attrib = get_file_attrib( filename );
        attrib = GetFileAttributesW( (LPCWSTR)filename );
     
       /* if( attrib == -1 )
            switch( errno )
            {
                case ENOENT:printf( "Маршрут или имя файла не найденыn" );
                    break;
                case EACCES:printf( "Отказ доступаn" );
                    break;
                default:    printf( "Ошибка номер %dn", errno );
                    break;
            }
        else*/
        {
            if( attrib & FILE_ATTRIBUTE_READONLY )
                printf( "%s имеет атрибут только на чтениеn", filename );
            if( attrib & FILE_ATTRIBUTE_HIDDEN )
                printf( "%s - cкрытый файлn", filename );
            if( attrib & FILE_ATTRIBUTE_SYSTEM )
                printf( "%s - системный файл n", filename );
           // if( attrib & FA_LABEL )
           //     printf( "%s - метка томаn", filename );
            if( attrib & FILE_ATTRIBUTE_DIRECTORY )
                printf( "%s - каталогn", filename );
            if( attrib & FILE_ATTRIBUTE_ARCHIVE )
                printf( "%s - архивный файлn", filename );
        }
     
        return 0;
    }

    Код

    Введите имя файла:test.txt
    test.txt имеет атрибут только на чтение
    test.txt - cкрытый файл
    test.txt - системный файл
    test.txt - каталог
    test.txt - архивный файл
    
    Process returned 0 (0x0)   execution time : 39.208 s

    Добавлено через 1 минуту
    ASORGOO, напильником пошоркай, если что…



    1



    0 / 0 / 0

    Регистрация: 11.01.2020

    Сообщений: 47

    29.05.2021, 19:04

     [ТС]

    5

    Весьма благодарю вас



    0



    1. 07-02-2013


      #1

      critixal is offline


      Registered User


      Question ‘a’ undeclared (first use in this function) Error

      Hello Programmers, I am trying to make a program that will allow a user to enter their name, from this point, three options should be present for the user to choose, then from that point, two more options should be present. I have if else statements in a switch case and I get the «undeclared» error for the a in the first » if(specificage == a) «. Does anyone know the fix to this, it is the only error I get. I’m new to programming, sorry if I sound like a rookie. Run it if you need, thanks!

      NOTE: I use Code::Blocks.

      Here is the code:

      Code:

      #include <stdio.h>
      #include <string.h>
      
      
      int main()
      {
          char name[50];
          char ageclass[50];
          char specificage[50];
      
      
          printf("Please enter your name: ");
          fgets(name, 50, stdin);
          printf("Hello %s", name);
      
      
          printf("Are you a:n");
          printf("1. Minorn");
          printf("2. Adultn");
          printf("3. Senior Citizenn");
          printf("Answer(1, 2, or 3): ");
          fgets(ageclass, 50, stdin);
      
      
          switch (ageclass[50])
          {
              case'1':
              {
                  printf("You are between the ages of:n");
                  printf("a. 0-12n");
                  printf("b. 13-18n");
                  printf("Answer(a or b): ");
                  fgets(specificage, 50, stdin);
                     if(specificage == a)
                      {
                          printf("You are a young minor.");
                      }
                      else
                      {
                          printf("You are an older minor.");
                      }
              }
      
      
              case'2':
              {
                  printf("You are between the ages of:n");
                  printf("a. 19-50n");
                  printf("b. 51-65n");
                  printf("Answer(a or b): ");
                  fgets(specificage, 50, stdin);
                      if(specificage == a)
                      {
                          printf("You are a young adult.");
                      }
                      else
                      {
                          printf("You are an older adult.");
                      }
              }
      
      
              case'3':
              {
                  printf("You are between the ages of:n");
                  printf("a. 66-90n");
                  printf("b. 91-110n");
                  printf("Answer(a or b): ");
                  fgets(specificage, 50, stdin);
                      if(specificage == a)
                      {
                          printf("You are a young senior citizen.");
                      }
                      else
                      {
                          printf("You are an older senior citizen.");
                      }
              }
          }
      
      
          getchar();
          return 0;
      }

      Last edited by critixal; 07-02-2013 at 05:49 PM.


    2. 07-02-2013


      #2

      critixal is offline


      Registered User


      Don’t reply, issue resolved!


    3. 07-02-2013


      #3

      anduril462 is offline


      Registered User


      You have two problems. First

      Code:

      if(specificage == a)

      a here is treated as a variable. If you want to check that the user input the letter a, you need to put single quotes around it: ‘a’. If you want a string containing one character, the letter a, use double quotes: «a»

      Second, specificage is a char array/string. You can’t compare strings with ==. You can compare individual characters in the string/array with ==, or use strcmp/strncmp:

      Code:

      if (strcmp(specificage, "some string") == 0)  // strcmp returns 0 if they're equal
      // or
      if (specificage[3] == 'q')  // check that the 4th (remember, arrays in C start at 0) character is a 'q'

      I’m guessing you were more interested in the second example.

      Two more notes:
      1. fgets leaves the newline character (remember, the user pressed enter after ‘a’ or ‘b’). If I enter «foo» followed by enter, the string fgets gives back contains «foon». If you do strcmp, it will not match «foo», since the n is significant. If you need to get rid of it, use the strchr trick here: FAQ > Get a line of text from the user/keyboard (C) — Cprogramming.com.
      2. The user may enter uppercase ‘A’ or ‘B’ for some reason, your program should handle it. Easy enough, use the toupper() or tolower() functions before comparing (remember to #include <ctype.h>)


    4. 07-02-2013


      #4

      critixal is offline


      Registered User


      Quote Originally Posted by anduril462
      View Post

      You have two problems. First

      Code:

      if(specificage == a)

      a here is treated as a variable. If you want to check that the user input the letter a, you need to put single quotes around it: ‘a’. If you want a string containing one character, the letter a, use double quotes: «a»

      Second, specificage is a char array/string. You can’t compare strings with ==. You can compare individual characters in the string/array with ==, or use strcmp/strncmp:

      Code:

      if (strcmp(specificage, "some string") == 0)  // strcmp returns 0 if they're equal
      // or
      if (specificage[3] == 'q')  // check that the 4th (remember, arrays in C start at 0) character is a 'q'

      I’m guessing you were more interested in the second example.

      Two more notes:
      1. fgets leaves the newline character (remember, the user pressed enter after ‘a’ or ‘b’). If I enter «foo» followed by enter, the string fgets gives back contains «foon». If you do strcmp, it will not match «foo», since the n is significant. If you need to get rid of it, use the strchr trick here: FAQ > Get a line of text from the user/keyboard (C) — Cprogramming.com.
      2. The user may enter uppercase ‘A’ or ‘B’ for some reason, your program should handle it. Easy enough, use the toupper() or tolower() functions before comparing (remember to #include <ctype.h>)

      Oh, actually this helped a lot, I will make these changes, thank you tons!
      Could you check out my other post, the more recent one, thanks! It involves the same program.

      Last edited by critixal; 07-02-2013 at 06:33 PM.


     
    `variable' undeclared (first use in this function)
    In C and C++ variables must be declared before they can be used. This
    error message indicates that the compiler has encountered a variable
    name which does not have a corresponding declaration. It can be caused
    by a missing declaration, or a typing error in the name. Variable names
    are case-sensitive, so foo and Foo represent different
    variables. To keep the output short, only the first use of an
    undeclared variable is reported.
    Example:

    int
    main (void)
    {
      int i;
      j = 0;     /* undeclared */
      return j;
    }
    

    The variable j is not declared and will trigger the error `j'
    undeclared
    .

    parse error before `...'
    syntax error
    These error messages occur when the compiler encounters unexpected
    input, i.e. sequences of characters which do not follow the syntax of
    the language. The error messages can be triggered by a missing close
    bracket, brace or semicolon preceding the line of the error, or an
    invalid keyword.
    Example:

    #include <stdio.h>
    
    int
    main (void)
    {
      printf ("Hello ")     /* missing semicolon */
      printf ("World!n");
      return 0;
    }
    

    There is a missing semicolon after the first call to printf,
    giving a parse error.

    parse error at end of input
    This error occurs if the compiler encounters the end of a file
    unexpectedly, such as when it has parsed an unbalanced number of opening
    and closing braces. It is often caused by a missing closing brace
    somewhere.
    Example:

    #include <stdio.h>
    
    int
    main (void) 
    {
      if (1) {
        printf ("Hello World!n");
        return 0;  /* no closing brace */
    }
    

    An additional closing brace is needed in this program to prevent the
    error parse error at end of input.

    warning: implicit declaration of function `...'
    This warning is generated when a function is used without a prototype
    being declared. It can be caused by failing to include a header file,
    or otherwise forgetting to provide a function prototype.
    Example:

    int
    main (void)
    {
      printf ("Hello World!n");  /* no header */
      return 0;
    }
    

    The system header file ‘stdio.h’ is not included, so the prototype
    for printf is not declared. The program needs an initial line
    #include <stdio.h>.

    unterminated string or character constant
    This error is caused by an opening string or character quote which does
    not have a corresponding closing quote. Quotes must occur in matching
    pairs, either single quotes 'a' for characters or double quotes
    "aaa" for strings.
    Example:

    #include <stdio.h>
    
    int
    main (void)
    {
      printf ("Hello World!n);  /* no closing quote */
      return 0;
    }
    

    The opening quote for the string in this program does not have a
    corresponding closing quote, so the compiler will read the rest of the
    file as part of the string.

    character constant too long
    In C and C++ character codes are written using single quotes, e.g.
    'a' gives the ASCII code for the letter a (67), and 'n'
    gives the ASCII code for newline (10). This error occurs if single quotes
    are used to enclose more than one character.
    Example:

    #include <stdio.h>
    
    int
    main (void)
    {
      printf ('Hello World!n');  /* wrong quotes */
      return 0;
    }
    

    The program above confuses single-quotes and double-quotes. A sequence
    of characters should be written with double quotes, e.g. "Hello
    World!"
    . This same problem occurs in the following C++ program,

    #include <iostream>
    
    int
    main (void)
    {
      std::cout << 'Hello World!n';  // wrong quotes
      return 0;
    }
    

    This error can also occur if the forward slash and backslash are
    confused in an escape sequence, e.g. using '/n' instead of
    'n'. The sequence /n consists of two separate
    characters, ‘/’ and ‘n’.
    Note that according to the C standard there is no limit on the length of
    a character constant, but the value of a character constant that
    contains more than one character is implementation-defined. Recent
    versions of GCC provide support multi-byte character constants, and
    instead of an error the warnings multiple-character character
    constant
    or warning: character constant too long for its type
    are generated in this case.

    warning: initialization makes integer from pointer without a cast
    This error indicates a misuse of a pointer in an integer context.
    Technically, it is possible to convert between integer and pointer
    types, but this is rarely needed outside system-level applications.
    More often, this warning is the result of using a pointer without
    dereferencing it (e.g. writing int i = p instead of int i
    = *p
    ).
    This warning can also occur with char and char * types,
    since char is an integer type.
    Example:

    int
    main (void)
    {
      char c = "n";  /* incorrect */
      return 0;
    }
    

    The variable c has type char, while the string "n"
    evaluates to a const char * pointer (to a 2-byte region of memory
    containing the ASCII value for newline followed by a zero byte '',
    since strings are null-terminated). The ASCII code for newline can be
    found using char c = 'n';
    Similar errors can occur with misuse of the macro NULL,

    #include <stdlib.h>
    
    int
    main (void)
    {
      int i = NULL;  /* incorrect */
      return 0;
    }
    

    In C, the macro NULL is defined as ((void *)0) in
    ‘stdlib.h’ and should only be used in a pointer context.

    dereferencing pointer to incomplete type
    This error occurs when a program attempts to access the elements of
    struct through a pointer without the layout of the struct being declared
    first. In C and C++ it is possible to declare pointers to structs
    before declaring their struct layout, provided the pointers are not
    dereferenced—this is known as forward declaration.
    Example:

    struct btree * data;
    
    int 
    main (void)
    {
      data->size = 0;  /* incomplete type */
      return 0;
    }
    

    This program has a forward declaration of the btree struct
    data. However, the definition of the struct is needed before the
    pointer can be dereferenced to access individual members.

    warning: unknown escape sequence `...'
    This error is caused by an incorrect use of the escape character in a
    string. Valid escape sequences are:

    n newline t tab
    b backspace r carriage return
    f form feed v vertical tab
    a alert (bell)

    The combinations \, ', " and ? can be
    used for individual characters. Escape sequences can also use octal
    codes 377 and hexadecimal codes
    x00xFF.
    Example:

    #include <stdio.h>
    
    int
    main (void)
    {
      printf ("HELLO WORLD!N");
      return 0;
    }
    

    The escape sequence N in the program above is invalid—the
    correct escape sequence for a newline is n.

    warning: suggest parentheses around assignment used as truth value
    This warning highlights a potentially serious error, using the assignment
    operator ‘=’ instead of the comparison operator ‘==’ in the
    test of a conditional statement or other logical expression. While the
    assignment operator can be used as part of a logical value, this is rarely
    the intended behavior.
    Example:

    #include <stdio.h>
    
    int
    main (void)
    {
      int i = 0;
      if (i = 1) {  /* = should be == */
        printf ("unexpected resultn");
      }
      return 0;
    }
    

    The test above should be written as if (i == 1), otherwise the
    variable i will be set to 1 by the evaluation of the if
    statement itself. The operator ‘=’ both assigns and returns the
    value of its right-hand side, causing the variable i to be
    modified and the unexpected branch taken. Similar unexpected results
    occur with if (i = 0) instead of if (i == 0),
    except that in this case the body of the if statement would
    never be executed.
    This warning is suppressed if the assignment is enclosed in additional
    parentheses to indicate that it is being used legitimately.

    warning: control reaches end of non-void function
    A function which has been declared with a return type, such as
    int or double, should always have a return
    statement returning a value of the corresponding type at all
    possible end points—otherwise its return value is not well-defined.
    Functions declared void do not need return statements.
    Example:

    #include <stdio.h>
    
    int
    display (const char * str)
    {
      printf ("%sn", str);
    }
    

    The program above reaches the end of the display function, which has
    a return type of int, without a return statement. An
    additional line such as return 0; is needed.
    When using gcc the main function of a C program must
    return a value of type int (the exit status of the program). In
    C++ the return statement can be omitted from the main
    function—the return value of the C++ main function defaults
    to 0 if unspecified.

    warning: unused variable `...'
    warning: unused parameter `...'
    These warnings indicate that a variable has been declared as a local
    variable or in the parameters of a function, but has not been used
    anywhere. An unused variable can be the result of a programming error,
    such as accidentally using the name of a different variable in place of
    the intended one.
    Example:

    int
    foo (int k, char * p)
    {
      int i, j;
      j = k;
      return j;
    }
    

    In this program the variable i and the parameter p are
    never used. Note that unused variables are reported by -Wall,
    while unused parameters are only shown with -Wall -W.

    warning: passing arg of ... as ... due to prototype
    This warning occurs when a function is called with an argument of a
    different type from that specified in the prototype. The option
    -Wconversion is needed to enable this warning. See
    the description of -Wconversion in section 3.5 Additional warning options for an example.
    warning: assignment of read-only location
    warning: cast discards qualifiers from pointer target type
    warning: assignment discards qualifiers ...
    warning: initialization discards qualifiers ...
    warning: return discards qualifiers ...
    These warnings occur when a pointer is used incorrectly, violating a
    type qualifier such as const. Data accessed through a pointer
    marked as const should not be modified, and the pointer itself
    can only be assigned to other pointers that are also marked
    const.
    Example:

    char *
    f (const char *s)
    {
      *s = '';  /* assigns to read-only data */
      return s;   /* discards const */
    }
    

    This program attempts to modify constant data, and to discard the
    const property of the argument s in the return value.

    initializer element is not a constant
    In C, global variables can only be initialized with constants, such as
    numeric values, NULL or fixed strings. This error occurs if a
    non-constant value is used.
    Example:

    #include <stdio.h>
    
    FILE *stream = stdout;  /* not constant */
    int i = 10;
    int j = 2 * i;          /* not constant */
    
    int
    main (void)
    {
      fprintf (stream, "Hello World!n");
      return 0;
    }
    

    This program attempts to initialize two variables from other variables.
    In particular, the stream stdout is not required to be a constant
    by the C standard (although on some systems it is a constant). Note
    that non-constant initializers are allowed in C++.

    Ezoic

     
    • Forum
    • Beginners
    • Undeclared (First Use This Function)?

    Undeclared (First Use This Function)?

    I’m trying to write a program that asks for the x and y coordinates of a point, then outputs which quadrant that point would be located under. However, my variable int qrant is not working in my functions. I receive the error message «‘qrant’ undeclared (first use this function)» in the main part of my program. Any help — even not related to this exact problem — would be appreciated. Thanks.

    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    13
    14
    15
    16
    17
    18
    19
    20
    21
    22
    23
    24
    25
    26
    27
    28
    29
    30
    31
    32
    33
    34
    35
    36
    37
    38
    39
    40
    41
    42
    43
    44
    45
    46
    47
    48
    49
    50
    51
    52
    53
    54
    55
    56
    57
    58
    59
    60
    61
    62
    63
    64
    65
    66
      #include <iostream>
    using namespace std;
    
    void inp(int& x_cord, int& y_cord);//sets 'x_cord' and 'y_cord' equal to user's input
    void calc(int& x_cord, int& y_cord, int& qrant);//checks the values of the x and y coordinates and finds the corresponding quadrant
    void outp(int& x_cord, int& y_cord, int& qrant);//outputs results in terms of quadrant or axis intercept
    
    int main()
    {
        char cont = 'y';
        int x_cord, y_cord;
        
        do//do-while loop allows program to be run multiple times
        {inp(x_cord, y_cord);
        calc(x_cord, y_cord, qrant);
        outp(x_cord, y_cord, qrant);
        cin >> cont;}
        while(cont == 'y' || cont == 'Y');
        
        return 0;}
        
    void inp(int& x_cord, int& y_cord)//***INPUT***
    {
        cout << "Please enter the x and y coordinates of a point on a graph, and I will tell you what quadrant that point is in.n"//prompt
             << "Enter the x coord (whole number): ";
        //Take 'x_cord'
        cin >> x_cord;
        cout << "Enter the y coord (whole number): ";
        //Take 'y_cord'
        cin >> y_cord;
        cout << endl << "The point you have given me is (" << x_cord << ", " << y_cord << ")nn";}
    
    void calc(int& x_cord, int& y_cord, int& qrant)//***CALCULATIONS***
    {
        if(x_cord > 0 && y_cord > 0)//I
        qrant = 1;
        if(x_cord < 0 && y_cord > 0)//II
        qrant = 2;
        if(x_cord < 0 && y_cord < 0)//III
        qrant = 3;
        if(x_cord > 0 && y_cord < 0)//VI
        qrant = 4;}
        
    void outp(int& x_cord, int& y_cord, int& qrant)//***OUTPUT***
    {
        //Y-AXIS
        if(x_cord == 0)
        cout << "Your point lies on the y-axis.";
        else
        {cout << "Your point is in Quadrant ";
        if(qrant == 1)
        cout << "I";
        if(qrant == 2)
        cout << "II";
        if(qrant == 3)
        cout << "III";
        if(qrant == 4)
        cout << "VI";}  
        //X-AXIS
        if(y_cord == 0)
        cout << "Your point lies on the x-axis.";
        //ORIGIN
        if(x_cord == 0 && y_cord == 0)
        cout << "Your point lies on the origin.";
        
        cout << endl << endl << endl <<  "Do you want to continue? < 'y' or 'n' >: ";}

    You never declared

    qrant

    in

    main()

    .

    Thanks. I had never seen the error message so I was really confused!

    Topic archived. No new replies allowed.

    Понравилась статья? Поделить с друзьями:
  • Und ошибка на машинке haier
  • Uncr data error cyclic redundancy check victoria
  • Uncorrectable pci express error embedded device bus 0 device 0 function 0 error status 0x00100000
  • Uncorrectable memory error processor 1 memory module 4
  • Uncorrectable errors count как исправить