I may be mistaken, but I'm think that $c+=1 is actually defined to run the two steps one right after the other, while what is ambiguous in $c++ is the fact that $c will be incremented later.

I do agree on the fact that undefined behaviour does not mean undefined value, that's why it's hard to track.

In reply to Re^5: Why does the first $c evaluate to the incremented value in [$c, $c += $_] ? by Eily
in thread Why does the first $c evaluate to the incremented value in [$c, $c += $_] ? by smls

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