in reply to Re^4: map in void context
in thread map in void context

It's not really returning "true", it's returning the value on the right side of the assignment. 

my $x = 2;
print "\$x is '$x'\n";

if ( $x = 0 ) {
    print "\$x = 0 is true\n";
}
else {
    print "\$x = 0 is false\n";
}

print "\$x is '$x'\n";
__END__
$x is '2'
$x = 0 is false
$x is '0'

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This behavior is why you can say "$x = $y = $foo" and set both $x and $y to whatever is in $foo.

If you consider the "effect" of assignment to be returning the value of the right hand side, then perhaps changing the value of the variable is a side effect. I don't see it that way, however.

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Re^6: map in void context
by ysth (Canon) on Dec 22, 2008 at 23:03 UTC
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Re^6: map in void context
by pat_mc (Pilgrim) on Dec 21, 2008 at 21:02 UTC
    Hi, kyle -

    Thanks for correcting my comment regarding the return value of the assignment operator =. I guess, the actual return value (namely the value assigned) is a perfect example for the informative return values of Perl operators I was commenting on earlier - and further down - in this thread.
    If you consider the "effect" of assignment to be returning the value of the right hand side, then perhaps changing the value of the variable is a side effect.
    I would say it exactly the opposite way: The effect of assignment is assigning a value. Its side-effect is that it also provides a return value (which just makes me wonder about list assignments ... must check if their return value really is a list of all assigned values).

    But OK - I think we now understand the way the other party is seeing things. Agreement must not necessarily be achieved. Thanks for responding again, anyway.
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[d/l]

      which just makes me wonder about list assignments ... must check if their return value really is a list of all assigned values

      In list context, list assignment evaluates to the list of elements actually assigned. 

      my @x = ($a,$b) = (4,5,6);   # @x = (4,5)

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      In scalar context, list assignment evaluates to the number of elements to assign. 

      my $x = ($a,$b) = (4,5,6);   # $x = 3

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      Remember, x = y = z means x = ( y = z ) because assignment is right-associative.

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        Thanks, ikegami, for the clear proof that indeed the side effect of list assignment is that a list of all values assigned is returned!

        This is a consistent extension of the behaviour of the assignment operator in scalar context and exactly in line with what I would have expected.

        To re-state my point: ikegami's code only works because of the side-effects associated with the assignment operator (To my personal surprise, ikegami mentioned earlier in this thread that
        Side-effects are something to be avoided when programming.
[reply]

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