That surprises me a little. I'd assume people would have more trouble with that because the addition and assignment statement appears to violate algebraic rules:

$a = $a + $b;

Obviously Larry's thought about that before, which is why we have:

$a += $b;

instead. Once they wrap their heads around the former, the inconsistency of chop trips them up again? Interesting.

In reply to RE: RE: RE: RE: (Ovid) RE(2): modify variable on pass by value by chromatic
in thread modify variable on pass by value by Anonymous Monk

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