[erkant]

I have the following lines of code in bash, and if I run it, my script shows an error in the while loop saying: intiger expression expected, and end the end the result of the execution is wrong.

#!/bin/bash

#
# Parity check
#

echo "Please enter a number: "
read num

if [ $num%2 -eq  0 ];
then
	echo "$num is even!"
else
	echo "$num is odd!"
fi

But if I change my if-statement to this:

if (( $num%2 == 0 ))

then it works nicely.

The point is that I get this error all the time. It doesn't matter whether it is if-statement or while-loop or similar things. I mean if I use the usual comparison keywords from bash and square brackets, it doesn't work. But if I use the other C based comparison signs, and parentheses it works nicely. Is it the problem that I don't use bash-like things correctly?

[sepp2k]

test/[ does not evaluate the operands as of the given operator as arithmetic expressions. It just sees the string 42%2 and notes that that string is not a valid number. You can expr to evaluate arithmetic expresions, so if [ $(expr $num % 2) -eq 0 ] works. Of course using double parentheses works, too, so you might as well keep using that.

[erkant]

Thank you very much sepp2k for your helpful and informative reply!