01 March 2009

I expected that the following script would print a error number. But it actually prints 0.

`echo hello world`;
$retval=`err-cmd >/dev/null 2>&1;echo $?`;
print ("result=$retval");

What is wrong? learning perl sec14.4 says:

The value beween backquotes is like the single-argument form of system and is interpreted as a doublequoted string, meaning that backslash-escapes and variables are expanded appropriately.

When perl interprets the aforementioned script, it will replace the $? with return value of the command before it. Since that value is 0 (the return value of echo hello world), the script is equivalent to this:

`echo hello world`;
$retval=`err-cmd >/dev/null 2>&1;echo 0`;
print ("result=$retval");

The fix is straight forward: escape the dollar sign:

$retval=`err-cmd >/dev/null 2>&1;echo \$?`;

Please note that shell scripts has a similar issue:

$ cat test.sh
eval "touch /file/not/exist 2>/dev/null; echo $?"
echo "$(touch /file/not/exist 2>/dev/null) $?"

$ ./test.sh

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