Perlsub talks about subroutine return values:

A "return" statement may be used to exit a subroutine, optionally specifying the returned value, which will be evaluated in the appropriate context (list, scalar, or void) depending on the context of the subroutine call. If you specify no return value, the subroutine returns an empty list in list context, the undefined value in scalar context, or nothing in void context. If you return one or more aggregates (arrays and hashes), these will be flattened together into one large indistinguishable list.

If no "return" is found and if the last statement is an expression, its value is returned. If the last statement is a loop control structure like a "foreach" or a "while", the returned value is unspecified. The empty sub returns the empty list.

Based on that, I thought I understood subroutine return values - but now I'm not sure anymore. Consider the following: 

sub test { 
  print "in test!";  
  return if 1 
}

if (&test) { 
  print "true" 
} else { 
  print "false" 
}

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Running this, I get: 
in test!
false
Which made sense to me - we're returning an empty list, which is false. Fine.

But I expected the behaviour to be flipped if I change the 1 to 0. Since 'return if 0' doesn't fire, we don't have an explicit return value, and so the sub should evaluate to the whatever the last statement that executed evaluated to - in this case, the print, which should be true.

But that wasn't what happened at all. I had to *comment out* the 'return' line to get the true value I was expecting.

So what am I missing here? Does the 'if' line count as a statement that evaluates to false?

-- zigdon

In reply to return if 0 by zigdon

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