comment on

Fun.

The reason that using your preInc() sub changes the result is because the value it returns is a copy of $m's value at the end of the call to the sub.

By contrast, the value of $m used by the + operator in the non-sub statement is the value of $m at the point after both the subexpressions involved in the + operation have been evaluated.

I assume that the order of the evaluation is a result of converting the expressions to their reverse polish form. Clear as mud, but this is (roughly) equivalent

$m=20; 
$r = \$m; 
print ${ $$r += 1; $r } + ${ my $t=$$r; $$r += 1; \$t };
43
[download]

At least, it produces the same result and is close enough for my mental processes to get a notional handle on the mechanism.

Examine what is said, not who speaks.

"Efficiency is intelligent laziness." -David Dunham
"Think for yourself!" - Abigail
"Memory, processor, disk in that order on the hardware side. Algorithm, algoritm, algorithm on the code side." - tachyon

In reply to Re: Quantum Weirdness and the Increment Operator by BrowserUk
in thread Quantum Weirdness and the Increment Operator by barrachois

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