comment on

That's not quite true. Observe:

# Does not say "true". This is as you describe.
say "true" if undef, undef;

# Still doesn't say true.
say "true" if 1, 2, 3, 4, undef;
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The reason for this behaviour is that if imposes a scalar context on the expression that acts as its condition. When the comma operator is used in a scalar context it is not a list constructor and instead evaluates the left argument, discards it and then evaluates the right argument (it basically acts like the semicolon but with a different precedence).

Thus this:

say "true" if 1, 2, 3, 4, undef;
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Is essentially the same as:

sub x {
  1;
  2;
  3;
  4;
  return undef;
}
say "true" if x();
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perl -E'sub Monkey::do{say$_,for@_,do{($monkey=[caller(0)]->[3])=~s{::}{ }and$monkey}}"Monkey say"->Monkey::do'

In reply to Re^3: What is true and false in Perl? by tobyink
in thread What is true and false in Perl? by root

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