Question

At work, we encountered an error when interacting with child processes in an object’s destructor, and eventually traced it to the $? variable being overwritten during the wait calls. This happens after the call to exit(), so $? additionally meant our program’s return code to the operating system.

Specifically, the perldoc talked about this sort of error:

Inside an END subroutine $? contains the value that is going to be given to exit(). You can modify $? in an END subroutine to change the exit status of your program.

We don’t want that to happen, so we put a local $?=$?; inside of every END block. But now the programs return success to the OS while actually failing in their given task.

I managed to break it down into two sample programs. One that works as intended, and one that fails. This occurs on both v5.8.8 and v5.10.1 for x86_64-linux-thread-multi

Program A: (returns 0 to the operating system)

END{ local $?=$?; }
exit(100);

Program B: (returns 100 to the operating system)

END{ local $?=$?>>8; }
exit(100);

Why does it matter what value was assigned to the local $? in the end block?

Answer 1

Looks like a bug in perl. Apparently self assignment of $? in local is broken:

% perl -wle '$? = 123; print "before: $?"; local $? = $?; print "after: $?"'    
before: 123
after: 0

But this version works fine:

% perl -wle '$? = 123; print "before: $?"; local $? = $? + 0; print "after: $?"'
before: 123
after: 123

Pretty bizarre.

A bug report has been filed.