Syntax error invalid arithmetic operator error token is

«invalid arithmetic operator» in shell

then I modify the test.sh:

/bin/bash -x test.sh

why this error appears, any suggestion will be appreciate!

3 Answers 3

only appears to work. Since arr is an ordinary array, not an associative array, it can only be indexed by non-negative integer values.

Consider this valid code:

All the echo statements print val . Since the index is treated as an arithmetic expression, it can refer to a variable (in this case, $index ) either with or without the $ prefix — even if it’s a quoted string.

In your code, where you never assigned a value to $index , $]> expands to $ , which is equivalent to $ , which is treated (by default) as equivalent to $ .

(If you have set -o nounset , then references to unset variables are treated as errors, and your code will produce an error message.)

Your second chunk of code:

is invalid because index.index is not a valid variable name — even though you probably meant it to be just a string used as an array index.

If you want arr to permit arbitrary strings as indices, it needs to be an associative array. You can create a non-associative array simply by assigning to it (or by using declare -a ), but an associative array can only be created with declare -A .

Associative arrays were added to bash in version 4. If you’re using an earlier version of bash, declare -A is not supported. You’ll need to upgrade to a newer bash, code up some clumsy alternative, or use a language that does support associative arrays, like Awk, Python, or Perl.

Adding declare -A arr (as user000001’s answer suggests) should solve the problem (if you have bash 4), but it’s instructive to understand what your original code is actually doing (or rather not doing).

(BTW, thanks for asking this; I learned a lot as I was composing this answer.)

Источник

bash — syntax error invalid arithmetic operator error token is

Приветствую! Ребята подскажите где я не догоняю?

Нужно выполнить проверку значений с первого массива во втором.

Элементы массива закавычь

Пробовал и ‘ кавычить и » кавычить. Одно и тоже.

Это из-за противоестественного интеллекта такое, если цифры — значить индекс, а не ассоциация. Добавьте везде букву при заполнении и сравнениях.

Жесть, а нет может ключа какого? что бы указал что здесь строка арифметика не нужна.

Добавил слово ip /usr/net/net-up: line 67: ip172.20.0.3/24: syntax error: invalid arithmetic operator (error token is «.20.0.3/24»)

Нате. Специально добавил несуществующий в конец current_ifip

На bash’e любой !Ъ сможет, даешь POSIX shell!

state_ip это не ассоциативный массив вообще-то, не смущает?

У вас тоже не bourne 🙂 Все эти $(cmd) $

. Если уж юзаете set —, то накой вообще этот load()? Делайте IFS=» » set —. Ну и т д.

А это башизм, да 🙂

Если уж юзаете set —, то накой вообще этот load()?

Чисто эстетически смотрится лучше, ИМХО. Сам смысл load() что бы построить подобие хеш-таблицы.

Этого не понял. Зачем IFS менять в данном случае ? .

Да это чисто новомодный эфемизм к обратным апострофам, мож и POSIX, но не такой уж и древний.

Для гарантии. Это правильный стиль, ибо меняет не глобально и самодокументированно для демонстрации чего хотят сделать в этой строчке.

Это правильный стиль, ибо меняет не глобально и самодокументированно для демонстрации чего хотят сделать в этой строчке.

Я понимаю, что он делает, и для чего нужен в целом. Я спрашивал за другое — чем или как именно замена IFS поможет конкретно в этом случае ? .

чем или как именно замена IFS поможет конкретно в этом случае

Сделает ровно то что вы делаете load() — распарсит строку по аргументам с разделителем пробел.

Почти то что нужно, но не совсем. Помогите доработать.

Первый раз ip добавляются второй раз прогоняю скрипт он снова пытается добавить. Т.е. не отрабатывает проверка в списке.

Источник

bash syntax error: invalid arithmetic operator with token «@mail.com»

I have a script that performs a SQL SELECT and sends some e-mails using the output but there is an issue with the script causing the error message:

line 34: johne@mail.com: syntax error: invalid arithmetic operator (error token is «@mail.com»)

Here is the script:

2 Answers 2

The problem is that you are trying to perform a string comparison inside an arithmetic expression. Either switch to a compound expression:

or check that you are accessing the correct element of msgData .

It first seemed to me that this is a limitation of the implementation of associative arrays in bash (as of v4.3.11 here).

Associative arrays are created using

Arrays are assigned to using compound assignments of the form

where each value is of the form [subscript]= string.

And then you look up subscript a few lines earlier:

The subscript is treated as an arithmetic expression that must evaluate to a number.

So it’s saying a string «foo@bar» can’t evaluate to a number, and can’t be used as an array key? But that doesn’t explain why strings without «@» can.

The confusing part is also that when you do the assignment manually, e.g.

That works fine. Bother.

The solution for this, as it turns out, is to repeat the declare -A part in the assignment, like this:

In the specific example above, it’s probable that $ gets evaluated in integer context because you use the comparison.

So I’d wager that you want to do something like quote it to escape that context first.

Or simply do it with fewer shorthands, without the :- modifier, which really contributes to the impression of line noise when combined with all the other punctuation on that line. For example:

Источник

getting syntax error: invalid arithmetic operator (error token is «. «) #357

This might not be a bash-completion issue, but I hope at least by explaining the issue that I could get some help debugging the issue.

The issue is this:

$ a.b bash: a.b: syntax error: invalid arithmetic operator (error token is «.b»)

If I source bash_completion like so:

everything works like expected:

In my .bashrc I have the following

I have also tried explicitly sourcing /usr/local/Cellar/bash-completion@2/2.8/share/bash-completion/bash_completion in my bashrc, but seems to have no effect. For some reason it only works when I do it in my shell.

The text was updated successfully, but these errors were encountered:

See «Troubleshooting» in README.md.

I’ve tried sourcing bash_completion also in my .bash_profile, also my .bash_profile only sources .bashrc so this should be ok.
I’ve tried to run the same commands with set +v and set +x , but not getting any more useful output.

Or was there any other sections in the Troubleshooting guide you would recommend?

To get a trace what’s happening when you hit tab, you should be running under e.g. set -x , not set +x (which turns it off).

Thanks. When running $ a.b [tab] now yields:

Ah, right, that’s fixed in commit a47bd37 which is included in 2.9.

. or probably not after all, actually. Anyway I’ll try to have a closer look soon.

I believe the error occurs when we try to access the _xspecs associative array to check if there’s a completion for a.b there. I can simulate the error for a non-associative array:

. but _xspecs is defined as an associative array, and I cannot reproduce with such one:

It’s like that in 2.8 too, so I have no more ideas at the moment.

I see. Like mentioned, this works when I explicitly source bash_completion again in my shell, even though I do it in my .bashrc as well — which sounds wierd. Any way to verify that the sourcing operation done in .bashrc was successful?

bash —norc gives you a «clean» shell, you can do set -x there and invoke the same commands your shell does when it starts up. I’d first triple check that what you manually source in your shell is the same bash_completion your rcfiles do.

The issue was resolved when I moved the sourcing of bash-completion out of a function inside my .bashrc . (https://unix.stackexchange.com/questions/151889/why-does-bashs-source-command-behave-differently-when-called-from-a-function)
Thank very much for help with debugging this. (edit: don’t think that issue had the same problem)

Not sure if this is something that could be detected and prevented, I’ll leave it up to you to decide.

Источник

invalid arithmetic operator (error token is » «) with for loop

gives error solution.sh: line 2: ((: c

I am using BASH. What is wrong with the for loop?

edit: I am working on the BASH hackerrank IDE and although this code is not directly related to the problem in this link, I am getting this error.

3 Answers 3

I’ve reproduced this error message by pressing Ctrl-E after 1. It looked like this:

So make sure you are not pressing some strange combination of keys before enter.

You need to add this line

at the top of solution.sh .

(if your bash is at a different location, do, in a terminal,

to determine its location)

@picasso13 just a wild guess cause it got me(and yield the same mysterious error when I tried to loop with array constructed from the input). There are 2 inputs on hackerRank(the first one is actually the size of the second). It solved my problem when I threw away the first and make sure my iteration was working on the list of numbers:

if you comment out my read ignore you’ll reproduce the issue.

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Introduction

Math and arithmetic operations are essential in Bash scripting. Various automation tasks require basic arithmetic operations, such as converting the CPU temperature to Fahrenheit. Implementing math operations in Bash is simple and very easy to learn.

This guide teaches you how to do basic math in Bash in various ways.

Prerequisites

• Access to the command line/terminal.
• A text editor to code examples, such as nano or Vi/Vim.
• Basic knowledge of Bash scripting.

Why Do You Need Math in Bash Scripting?

Although math is not the primary purpose of Bash scripting, knowing how to do essential calculations is helpful for various use cases.

Common use cases include:

• Adding/subtracting/multiplying/dividing numbers.
• Rounding numbers.
• Incrementing and decrementing numbers.
• Converting units.
• Floating-point calculations.
• Finding percentages.
• Working with different number bases (binary, octal, or hexadecimal).

Depending on the automation task, basic math and arithmetic in Bash scripting help perform a quick calculation, yielding immediate results in the desired format.

Bash Math Commands and Methods

Some Linux commands allow performing basic and advanced calculations immediately. This section shows basic math examples with each method.

Arithmetic Expansion

The preferable way to do math in Bash is to use shell arithmetic expansion. The built-in capability evaluates math expressions and returns the result. The syntax for arithmetic expansions is:

$((expression))

The syntax consists of:

• Compound notation (()) which evaluates the expression.
• The variable operator $ to store the result.

For example, add two numbers and echo the result:

echo $((2+3))

The arithmetic expansion notation is the preferred method when working with Bash scripts. The notation is often seen together with if statements and for loops in Bash.

awk Command

The awk command acts as a selector for pattern expressions. For example, to perform addition using the awk command, use the following example statement:

awk 'BEGIN { x = 2; y = 3; print "x + y = "(x+y) }'

For variables x = 2 and y = 3, the output prints x + y = 5 to the console.

bc Command

The bc command (short for basic calculator) is a command-line utility that renders the bc language. The program runs as an interactive program or takes standard input to perform arbitrary precision arithmetic.

Pipe an equation from standard input into the command to fetch results. For example:

echo "2+3" | bc

The output prints the calculation result.

dc Command

The dc command (short for desk calculator) is a calculator utility that supports reverse Polish notation. The program takes standard input and supports unlimited precision arithmetic.

Pipe a standard input equation into the command to fetch the result. For example:

echo "2 3 + p" | dc

The p in the equation sends the print signal to the dc command.

declare Command

The Bash declare command allows integer calculations. To use declare for calculations, add the -i option. For example:

declare -i x=2 y=3 z=x+y

Echo each variable to see the results:

echo $x + $y = $z

The output prints each variable to the console.

expr Command

The expr command is a legacy command line utility for evaluating integer arithmetic. An example expr command looks like the following:

expr 2 + 3

Separate numbers and the operation sign with spaces and run the command to see the calculation result.

factor Command

The factor command is a command-line utility that prints the factors for any positive integer, and the result factorizes into prime numbers.

For example, to print the factors of the number 100, run:

factor 100

The output prints the factored number.

let Command

The Bash let command performs various arithmetic, bitwise and logical operations. The built-in command works only with integers. The following example demonstrates the let command syntax:

let x=2+3 | echo $x

The output prints the results.

test Command

The test command in Linux evaluates conditional expressions and often pairs with the Bash if statement. There are two variations for the test syntax:

test 2 -gt 3; echo $?

Or alternatively:

[ 2 -gt 3 ]; echo $?

The test command evaluates whether two is greater than (-gt) three. If the expression is true, the output is zero (0), or one (1) if false.

Bash Arithmetic Operators

Bash offers a wide range of arithmetic operators for various calculations and evaluations. The operators work with the let, declare, and arithmetic expansion.

Below is a quick reference table that describes Bash arithmetic operators and their functionality.

How to Do Math in Bash

Bash offers different ways to perform math calculations depending on the type of problem.

Below are examples of some common problems which use Bash math functionalities or commands as a solution. Most examples use the Bash arithmetic expansion notation. The section also covers common Bash math errors and how to resolve them.

Math with Integers

The arithmetic expansion notation is the simplest to use and manipulate with when working with integers. For example, create an expression with variables and calculate the result immediately:

echo $((x=2, y=3, x+y))

To evaluate multiple expressions, use compound notation, store each calculation in a variable, and echo the result. For example:

((x=2, y=3, a=x+y, b=x*y, c=x**y)); echo $a, $b, $c

When trying to divide, keep the following in mind:

1. Division by zero (0) is impossible and throws an error.

2. Bash arithmetic expansion does not support floating-point arithmetic. When attempting to divide in this case, the output shows zero (0).

The result of integer division must be an integer.

Incrementing and Decrementing

Bash arithmetic expansion uses C-style integer incrementing and decrementing. The operator for incrementing or decrementing is either before or after the variable, yielding different behavior.

If the operator is before the variable (++x or --x), the increment or decrement happens before value assignment. To see how pre-incrementing works, run the following lines:

number=1

echo $((++number))

The variable increments, and the new value is immediately available.

If the operator is after the variable (x++ or x--), the increment or decrement happens after value assignment. To see how post-incrementing works, run the following:

number=1

echo $((number++))

echo $number

The variable stays the same and increments in the following use.

Floating-point Arithmetic

Although Bash arithmetic expansion does not support floating-point arithmetic, there are other ways to perform such calculations. Below are four examples using commands or programming languages available on most Linux systems.

1. Using awk for up to 6 decimal places:

awk 'BEGIN { x = 2.3; y = 3.2; print "x * y = "(x * y) }'

2. Using bc with the -l flag for up to 20 decimal places:

echo "2.3 * 3.2" | bc -l

3. Using Perl for up to 20 decimal places:

perl -e 'print 2.3*3.2'

Perl often comes preinstalled in Linux systems.

4. Using printf and arithmetic expansion to convert a fraction to a decimal:

printf %.<precision>f "$((10**<multiplier> * <fraction>))e-<multiplier>"

Precision dictates how many decimal places, whereas the multiplier is a power of ten. The number should be lower than the multiplier. Otherwise, the formula puts trailing zeros in the result.

For example, convert 1/3 to a decimal with precision two:

printf %.2f "$((10**3 * 1/3))e-3"

Avoid this method for precise calculations and use it only for a small number of decimal places.

Calculating a Percentage and Rounding

Below are two ways to calculate a percentage in Bash.

1. Use printf with arithmetic expansion.

printf %.2f "$((10**4 * part/total))e-4"%

For example, calculate what percent 40 is from 71:

printf %.2f%% "$((10**4 * 40/71))e-4"%

The precision is limited to two decimal places, and the answer always rounds down.

2. Use awk with printf for better precision:

awk 'BEGIN { printf "%.2f%%", (part/total*100) }'

For instance, calculate how many percent is 40 from 71 with:

awk 'BEGIN { printf "%.2f%%", (40/71*100) }'

The answer rounds up if the third decimal place is higher than five, providing better accuracy.

Finding a Factorial in the Shell

To calculate a factorial for any number, use a recursive Bash function.

For small numbers, Bash arithmetic expansion works well:

factorial () { 
    if (($1 > 1))
    then
        echo $(( $( factorial $(($1 - 1)) ) * $1 ))
    else

        echo 1
        return
    fi
}

To check the factorial for a number, use the following syntax:

factorial 5

The method is slow and has limited precision (up to factorial 20).

For higher precision, faster results, and larger numbers, use the bc command. For example:

echo 'define factorial(x) {if (x>1){return x*factorial(x-1)};return 1}
 factorial(<number>)' | bc

Replace <number> with the factorial number to calculate. For example, to find the factorial of 50, use:

echo 'define factorial(x) {if (x>1){return x*factorial(x-1)};return 1} factorial(50)' | bc

The output prints the calculation result to the terminal.

Creating a Bash Calculator Function

Create a simple Bash calculator function with the following code:

calculate() { printf "%sn" "[email protected]" | bc -l; }

The function takes user input and pipes the equation into the bc command.

Alternatively, to avoid using programs, use Bash arithmetic expansion in a function:

calculate() { echo $(("[email protected]")); }

Keep the arithmetic expansion limitations in mind. Floating-point arithmetic is not available with this function.

Save the function into the .bashrc file to always have the function available in the shell.

Using Different Arithmetic Bases

By default, Bash arithmetic expansion uses base ten numbers. To change the number base, use the following format:

base#number

Where base is any integer between two and 64.

For example, to do a binary (base 2) calculation, use:

echo $((2#1010+2#1010))

Octal (base calculations use a 0 prefix as an alias. For example:

echo $((010+010))

Hexadecimal (base 16) calculations allow using 0x as a base prefix. For example:

echo $((0xA+0xA))

The output prints the result in base ten for any calculation.

Convert Units

Create a simple Bash script to convert units:

1. Open a text editor, such as Vim, and create a convert.sh script. For example:

vim convert.sh

2. Paste the following code:

#!/bin/bash

## Program for feet and inches conversion

echo "Enter a number to be converted:"

read number

echo $number feet to inches:
echo "$number*12" | bc -l

echo $number inches to feet:
echo "$number/12" | bc -l

The program uses Bash read to take user input and calculates the conversion from feet to inches and from inches to feet.

3. Save the script and close:

:wq

4. Run the Bash script with:

. convert.sh

Enter a number and see the result. For different conversions, use appropriate conversion formulas.

Solving «bash error: value too great for base»

When working with different number bases, stay within the number base limits. For example, binary numbers use 0 and 1 to define numbers:

echo $((2#2+2#2))

Attempting to use 2#2 as a number outputs an error:

bash: 2#2: value too great for base (error token is "2#2")

The number is not the correct format for binary use. To resolve the error, convert the number to binary to perform the calculation correctly:

echo $((2#10+2#10))

The binary number 10 is 2 in base ten.

Solving «syntax error: invalid arithmetic operator»

The Bash arithmetic expansion notation only works for integer calculations. Attempt to add two floating-point numbers, for example:

echo $((2.1+2.1))

The command prints an error:

bash: 2.1+2.1: syntax error: invalid arithmetic operator (error token is ".1+2.1")

To resolve the error, use regular integer arithmetic or a different method to calculate the equation.

Solving «bash error: integer expression expected»

When comparing two numbers, the test command requires integers. For example, try the following command:

[ 1 -gt 1.5 ]

The output prints an error:

bash: [: 1.5: integer expression expected

Resolve the error by comparing integer values.

Conclusion

You know how to do Bash arithmetic and various calculations through Bash scripting.

For more advanced scientific calculations, use Python together with SciPy, NumPy, and other libraries.