Local ($?) In END blocks - Why is it important what value was assigned to it?

Question

At work, we encountered an error when interacting with child processes in the object's destructor and, ultimately, traced it to $? the variable will be overwritten during wait calls. This happens after calling exit (), so $? additionally means that our return code to the operating system.

In particular, perldoc talked about this error:

Inside the END $ routine? contains the value that will be returned by exit (). Can you change $? in the END routine, change the exit status of your program.

We do not want this to happen, so we put local $?=$?; inside each END block. But now programs return the success of the OS, without actually coping with their task.

I managed to break it into two test programs. One that works as intended, and one that fails. This happens 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 local $? in the final block?

Answer 1

Looks like a bug in perl. $? Self-determination apparently violated $? in local :

 % 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 strange.

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