Shell Programming and Scripting

Originally Posted by Scrutinizer

Aia, could you elaborate on why that should render a different outcome ?

In awk:

Code:

awk 'BEGIN {print (-2)%5, -2%5}'
-2 -2

Binary "%" is the modulo operator, which computes the division remainder of its first argument with respect to its second argument. Given integer operands $m and $n : If $n is positive, then $m % $n is $m minus the largest multiple of $n less than or equal to $m . If $n is negative, then $m % $n is $m minus the smallest multiple of $n that is not less than $m (that is, the result will be less than or equal to zero). If the operands $m and $n are floating point values and the absolute value of $n (that is abs($n)) is less than (UV_MAX + 1) , only the integer portion of $m and $n will be used in the operation (Note: here UV_MAX means the maximum of the unsigned integer type). If the absolute value of the right operand (abs($n)) is greater than or equal to (UV_MAX + 1) , "%" computes the floating-point remainder $r in the equation ($r = $m - $i*$n) where $i is a certain integer that makes $r have the same sign as the right operand $n (not as the left operand $m like C function fmod() ) and the absolute value less than that of $n . Note that when use integer is in scope, "%" gives you direct access to the modulo operator as implemented by your C compiler. This operator is not as well defined for negative operands, but it will execute faster.

Originally Posted by jim mcnamara

Code:

echo "(9-7) % 5 = $(( $(( 9 - 7 )) % 5 )) (should be 2)."
# and
echo "(9-7) % 5 = $(( ( 9 - 7 ) % 5 )) (should be 2)."

Note the spaces.

Both work in bash. /bin/sh does not have to always equal bash, it should a modern POSIX shell. I do not know how /bin/sh evaluates on busybox.

If $n is positive, then $m % $n is $m minus the largest multiple of $n less than or equal to $m.