Error expression must have integral or unscoped enum type

c++ – Error: Expression must have integral or unscoped enum type

Your variable size is declared as: float size;

You cant use a floating point variable as the size of an array – it needs to be an integer value.

You could cast it to convert to an integer:

float *temp = new float[(int)size];

Your other problem is likely because youre writing outside of the bounds of the array:

   float *temp = new float[size];

    //Getting input from the user
    for (int x = 1; x <= size; x++){
        cout << Enter temperature  << x << : ;

        // cin >> temp[x];
        // This should be:
        cin >> temp[x - 1];
    }

Arrays are zero based in C++, so this is going to write beyond the end and never write the first element in your original code.

c++ – Error: Expression must have integral or unscoped enum type

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The programming error “expression must have integral or unscoped enum type” occurs when the user has not entered any integer, integral, or some other technical reasons, such as; pointers or operators were not used properly.

This article will guide its readers about this issue and give them all possible solutions. Let’s begin with the reasons why this error occurs!

The expression must have integral or unscoped enum type error occurs due to multiple reasons, such as the integer is missing, an operator is wrong, or the pointer is not defined. Other conditions that could lead to this condition include newly allocated clones being freed and syntax errors.

Along with these, the following conditions play an important role:

• Wrong usage of operator
• Char points at nothing
• New allocated Clone was freed
• size is declared as float size
• writing outside of the array’s bounds
• Syntax error

– Wrong Usage of Operator

This error will arise when a wrong operator is used while solving integers. Moreover, an error message will pop up during the program’s execution when a wrong operator is used with the floating point operands.

Below is an example of a program with the wrong operator in use.

For example:

Lon 3 = (lon 2. to Rad() + L + 4* Math.PI) % (3* Math.PI) – Math.PI;

In this example, the programmer has used the wrong operator. The modulo operator % cannot be used with floating point operands.

– Char Points at Nothing

When a char is made, it also needs a defined pointer. However, if the program does not have a defined pointer, this error message will occur on the screen during the program’s execution.

An error will occur in the example below if the programmer adds both strings together. This is because the copied texts did not have any defined pointers.

Example of the program:

In this program, the user has created a char named string 1, which points at nothing. Therefore, Char points at nothing. Moreover, the text “I love” was copied into nothing, as there was no pointer defined, thus, causing an error and undefined behavior. The same is the case with Char* String 2 as well.

– New Allocated Clone Was Not Freed

If the program is not coded properly, it is possible that during the execution process of the program, the newly allocated Clone will never be freed. This will cause an undefined behavior that gives an error message of “expression must have integral or unscoped enum type”.

For example:

Here, the operator += is wrong, or a memory leak. It uses operator = (Char* ), and while executing the program, it would never free the newly allocated Clone. Thus, the program ends with an error.

– Size Is Declared as Float Size

When the variable in a program is declared as (float size;), the program will show an error. A floating point variable cannot be used as the size of an array; there has to be “int” in it. For example, when the programmer is trying to find the size of the array by using (float *temp = new float[size];), they get the “expression must have integral or unscoped enum type.” error message.

Furthermore, when trying the exact same coding on Cuda, a similar error occurs, which states: Cuda error: expression must have integral or unscoped enum type.

– Writing Outside of the Array’s Bounds

This type of integral error also occurs when the programmer types the program outside the boundaries of an array. Check the following example to understand it more clearly:

For example:

Here, the issue is in the (// cin >> temp[g];) program line. Arrays are zero-based in C++ programming language, so this program will write beyond the end but never the first element in the original code. Instead, it will show an integral error.

– Syntax Error

Syntax errors occur when the programmer uses the wrong functions or the sequence of functions, and the program is wrong. These syntax errors can also be following errors related to wrong expressions, missing integers, etc. Some possible errors can be:

• The expression must have integral or unscoped enum type modulo.
• The expression must have integral or enum type switch case.
• The expression must have integral type switch 0/.

How To Fix the Expression Must Have Integral or Unscoped Enum Type?

You can fix the expression that must have an integral or unscoped enum type error message by using correct operators and ensuring the integers or (int) function is not missing from the program. You can also get rid of this error by avoiding writing outside the array’s boundary.

However, the following solutions will help you solve this error efficiently:

– Use the Correct Operator

To correct this error message, the programmer should ensure they are using the correct operations in the program. For example, if they use the “%” function instead of the “fmod()” function to calculate the remainder, an error will occur.

Moreover, the modulo operator % cannot be used with floating point operands. So, use the fmod() function instead to calculate the remainder for floating point values. Therefore, the correct way of using an operator in integer based program is given below in the example

For example:

lon3 = (lon2.toRad()+L+4*Math.PI) fmod (3*Math.PI) – Math.PI;

– Ensure the Char Is Pointing at a Function

The programmer has to make sure that the Char they have created uses a pointer that is defined. Otherwise, anything that is written in it will be of no use as it will not get identified by the program and will create an error instead. Thus, use the Char or String that has a defined pointer.

The programmer tries to add two strings together in the following example, but the Char does not have defined pointers. Therefore, when they do char 1+= char 2, they’re trying to add two char pointers together.

This means they’re trying to do Char* += Char*, which the program does not have a defined operator for. The one that is defined is for SuperString += Char*.

For Example:

– Avoid Writing Outside the Bounds of an Array

Be careful not to write the program outside the bounds of an array. Otherwise, the integral error will pop up on the screen. This means whatever the programmer defines in an array, they have to use that instead of adding something new that is out of the bounds of an array. Here’s an example:

For Example:

Conclusion

After reading this guide, the reader can understand all the possible reasons behind this error message and its various solutions. The key takeaways from this article are:

• In C++, the programmer should always allocate integral values when coding an expression.
• If Char does not point at anything in the C programming language, a new error will occur, such as: expression must have integral type in C.
• Ensure the correct use of operators used during coding.

You can now easily solve this error in their programs by using this article as a guide.

References

• https://www.codeproject.com/Questions/373048/Error-Expression-must-have-integral-or-enum-type
• https://www.reddit.com/r/cpp_questions/comments/qvkaeb/overloading_concatenate_operator_expression_must/
• https://stackoverflow.com/questions/21341254/error-expression-must-have-integral-or-unscoped-enum-type

Error expression must have integral or enum type

So far I have the following code:

, but I get the following error message:

IntelliSense: expession must have integral or unscoped enum type

three times in a row for lines 25, 27, and 29 and I don’t understand or know why?

In case the purpose does make sense here are the directions:

2.7: Ocean Levels
Assuming the ocean’s level is currently rising at about 1.5 millimeters per year, write a program that displays
• The number of millimeters higher than the current level that the ocean’s level will be in 5 years,
• The number of millimeters higher than the current level that the ocean’s level will be in 7 years,
• The number of millimeters higher than the current level that the ocean’s level will be in 10 years,

Output labels: Each value should be on a line by itself, preceded by the a label of the form:
In X years the ocean’s level will be higher by Y millimeters.
where X is the number of years (5, 7 or 10) and Y is the value you calculate.

Does any else have an idea?

This is probably not doing what you’re expecting it too. X will only hold one value, not three.

You’re trying to do an assignment to Y within the three output statements. I’m guessing you didn’t mean to try to overwrite the year which the user entered in Y. What are you trying to output here?

cout «The number of millimeters higher the current level of the ocean in 5 years is; » ‘.’

If you edit your post, highlight the code part, then click on the <> button in the Format palette at the right of the post, your code will format for the forum with line numbers etc.

What your saying is that I can’t assign X three values, but I can assign X to multiple assignments?

Also where am I trying to do an assignment to Y within the three output statements and overwrite the year, which the user entered in Y?

Output labels: Each value should be on a line by itself, preceded by the a label of the form:
In X years the ocean’s level will be higher by Y millimeters.
where X is the number of years (5, 7 or 10) and Y is the value you calculate.

Quit thinking of X and Y as variables. They are simply place holders in the text of the problem description to indicate where certain values should be displayed in the output. They don’t necessarily have any place in the code as variables.

You have defined Y as a string and X as a float. The expression Y = X * 1.5 is nonsensical to the compiler. If you’re using C++11, you could use Y = std::to_string(X * 1.5) to get it to compile, however the assignment to string isn’t necessary because our std::ostream handles the display of floating point values just fine:
cout «The number . is: » «.n» ;

I know it says expression must have integral or unscoped enum type, but what does that mean? Also I tried a scoped enum as follows, but it didn’t like it either:

You’re overwriting the value of Y here each time you make a new assignment. And yearValue won’t have a value yet since the user doesn’t input the year until after this code.

The code still has this assignment statement in the output stream. (assignment meaning the code is doing a calculation and trying to assign it to Y.)

I might do something like this — I’ve just put one option here for 5 years, but it would be similar for the other years. Using a constant variable instead of hardcoding the specific number makes it easier later on to change the time periods without having to search through the code to update in multiple places. This is a short program, so it’s not a big deal here, but just something to keep in mind.

You might also want to add in some validation to make sure that they’ve input a valid year.

I solved a few problems but aquired some more unless they already existed. My current source code is as follows:

Any idea how to solve the following errors:

35 IntelliSense: no operator «>>» matches these operands
operand types are: std::istream >> std::string () c:Usersdgrossi0914DocumentsVisual Studio 2013ProjectsProgram2Project5Project5Source.cpp 28 6 Project5
30 IntelliSense: expression must have integral or unscoped enum type c:Usersdgrossi0914DocumentsVisual Studio 2013ProjectsProgram2Project5Project5Source.cpp 19 27 Project5
31 IntelliSense: expression must have integral or unscoped enum type c:Usersdgrossi0914DocumentsVisual Studio 2013ProjectsProgram2Project5Project5Source.cpp 20 27 Project5
32 IntelliSense: expression must have integral or unscoped enum type c:Usersdgrossi0914DocumentsVisual Studio 2013ProjectsProgram2Project5Project5Source.cpp 21 27 Project5
33 IntelliSense: expression must have a constant value c:Usersdgrossi0914DocumentsVisual Studio 2013ProjectsProgram2Project5Project5Source.cpp 24 38 Project5
34 IntelliSense: expected a ‘>’ c:Usersdgrossi0914DocumentsVisual Studio 2013ProjectsProgram2Project5Project5Source.cpp 24 48 Project5
Error 9 error C4430: missing type specifier — int assumed. Note: C++ does not support default-int c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 24 1 Project5
Error 11 error C4430: missing type specifier — int assumed. Note: C++ does not support default-int c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 25 1 Project5
Error 13 error C4430: missing type specifier — int assumed. Note: C++ does not support default-int c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 26 1 Project5
Error 16 error C4430: missing type specifier — int assumed. Note: C++ does not support default-int c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 27 1 Project5
Error 19 error C4430: missing type specifier — int assumed. Note: C++ does not support default-int c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 28 1 Project5
Error 22 error C4430: missing type specifier — int assumed. Note: C++ does not support default-int c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 30 1 Project5
Error 25 error C4430: missing type specifier — int assumed. Note: C++ does not support default-int c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 31 1 Project5
Error 6 error C2365: ‘X3’ : redefinition; previous definition was ‘data variable’ c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 21 1 Project5
Error 5 error C2365: ‘X2’ : redefinition; previous definition was ‘data variable’ c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 21 1 Project5
Error 4 error C2365: ‘X1’ : redefinition; previous definition was ‘data variable’ c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 21 1 Project5
Error 1 error C2296: ‘+’ : illegal, left operand has type ‘std::string (__cdecl *)(void)’ c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 17 1 Project5
Error 2 error C2296: ‘+’ : illegal, left operand has type ‘std::string (__cdecl *)(void)’ c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 18 1 Project5
Error 3 error C2296: ‘+’ : illegal, left operand has type ‘std::string (__cdecl *)(void)’ c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 19 1 Project5
Error 10 error C2143: syntax error : missing ‘;’ before ‘>>’ c:usersdgrossi0914documentsvisual studio 2013projectsprogram2project5project5source.cpp 25 1 Project5
Error 8 error C2143: syntax error : missing ‘;’ before ‘

You’ve defined YearValue to be a function that returns a string on line 10.

But then you try to add YearValue as if it were a variable. But if YearValue is a string and X1 is a double, you can’t add those. Also — the user hasn’t entered a value for the year at this point, so Y1 is not going to end up having a valid value.

double Y1 = (YearValue + X1) * 1.5;

The program is run through line by line sequentially — a later line in the code is not going to look back to this definition at the beginning of main to recalculate what Y1 should be when you try to cout the value.

You don’t need the enum lines. You really don’t even need the Y variables. You can output the value of the calculation (year+increment) * 1.5 directly.

First, quit paying attention to Intellisense messages. They’re helpful if you know what you’re doing, but you clearly do not. Instead, pay attention to errors generated when you actually attempt to compile. (Use the Output tab as opposed to the Error List tab, and choose Show output from Build.)

You need to revisit your specification. There is no requirement for you to ask the user for a year.

Lines 10, 11 and 12 in your code declare functions which are never defined.
Lines 16, 17 and 18 attempt to do pointer arithmetic with a function pointer. This is an error.
On line 20 you attempt to re-define identifiers already defined in the current scope. This is an error.
On line 21 you attempt to do pointer arithmetic with a function pointer. This is an error.

You’re really over-thinking this.

You’ve defined YearValue to be a function that returns a string on line 10.

But then you try to add YearValue as if it were a variable. But if YearValue is a string and X1 is a double, you can’t add those. Also — the user hasn’t entered a value for the year at this point, so Y1 is not going to end up having a valid value.

double Y1 = (YearValue + X1) * 1.5;

The program is run through line by line sequentially — a later line in the code is not going to look back to this definition at the beginning of main
Edit & Run
to recalculate what Y1 should be when you try to cout the value.

You don’t need the enum lines. You really don’t even need the Y variables. You can output the value of the calculation (year+increment) * 1.5 directly.»

I know you said I don’t need the enum lines, but I couldn’t figure out a way to get it to work with it for the X variable. However, I did get it to work without a Y variable as follows:

This really isn’t a situation where I’d use enums.

See Cire’s post above for an even more simplified approach.

Источник

I’m trying to overload the += operator so that it will concatenate strings. In my test code, I’m getting an error message when I try to call the operator. It says «Expression must have integral or unscoped enum type.» I think this has to do with not using the proper operands, but as far as I can tell, my operands are the correct type. I’ve tried using a string literal for the rvalue as well and it didn’t work either.

Here is the test code:

void ConcatenateOperatorTests()
{
	// object/variable declarations
	char* String1 = 0;
	strcpy(String1, "I love ");

	char* String2 = 0;
	strcpy(String2, "Star Trek");

	// tests
	String1 += String2; // this line causes the error, with the squiggly under String2

	// print results
	cout << String1 << endl;
}

And here is the code to overload the += operator. I’m well aware that there could easily be other problems with my operator code, but I would prefer to figure those out for myself unless they’re relevant to this particular problem.

void SuperString::operator+=(const char* StringToAppend)
{
	long ToAppendLength = 0;
	long OriginalLength = 0;
	long NewLength = 0;
	long Index = 0;
	long Index2 = 0;
	char* Clone = 0;
	long Length = 0;

	OriginalLength = Length();
	ToAppendLength = strlen(StringToAppend);
	NewLength = OriginalLength + ToAppendLength;

	// null?
	if (StringToAppend != 0)
	{
		// no
		Clone = new char[NewLength + 1];
		strcpy_s(Clone, NewLength + 1, StringToAppend);
	}
	else
	{
		// yes, default to empty string
		Clone = new char[1];
		*(Clone + 0) = 0;
	}

	for (Index = 0; Index > OriginalLength && Index < NewLength; Index++)
	{
		Clone[Index] = StringToAppend[Index2];
		Index2++;
	}

	*this = *Clone;
}

Thanks for any help, I really appreciate it