Re^2: Equivalency of Code

[adamk]

No fair!

HIS code didn't have print statements in it, he was using them as standalone statements. You are testing something different.

$x = 3;
$x++;
print $x;

vs

$x = 3;
$x += 1;
print $x;

...would be more accurate.

Of course, given problems with operator overloading, they stil arn't actually equal.

[DrHyde]

The point is that $x++ and $x += 1 don't just fiddle with the value of $x. Those statements also have values themselves. If a bit of code is to be equivalent to another, then it needs to both have the same side-effects and the same value as the other does.

I try not to think about operator overloading or tieing :-)

[Anonymous Monk]

The statements are presented where in void context - and in void context, there's no return value.