There is one bash quirk I just learnt today and that caused me fair amount of debugging. In summary when you use the set -e option in bash, any function that returns a non-zero exit code will cause your script to terminate except that this does not apply when the function is inside an if statement – more accurately when the function forms the argument of an if clause. We can see this behavior in action below:
#!/bin/bash
set -e
function func {
if [ "$1" = "$2" ] ; then
return 0
else
return 1
fi
}
VAR1=foo
VAR2=bar
if func $VAR1 $VAR2 ; then
echo "func returned 0"
else
echo "func returned 1. set -e does not cause script to exit."
fi
SOME_VAR=$(func $VAR1 $VAR1)
echo "SOME_VAR = $SOME_VAR. the function returns 0 so execution continues till here."
SOME_OTHER_VAR=$(func $VAR1 $VAR2)
echo "we never get here and program exits beforehand since function returns 1 and set -e causes script to exit."
Here is output of running this script:
$ ./test.sh
func returned 1
SOME_VAR = . the function returns 0 so execution continues till here.
$ echo $?
1
siddharth's space 