Re: Re: Re: Mysterious for behavior

[Belgarion]

Post-increment does not yield a new lvalue, but it does have an lvalue, and does modify the variable. Update: I don't what I was thinking about there. Post-increment has an rvalue. (That's what happens when you try to post a message just before quitting time!) However, below is still semi-interesting. Consider:

print "$_ and $i\n" for ($i = 1, $i++, $i++);

__OUTPUT__
3 and 3
1 and 3
2 and 3

The $i is being modified, and the value before the post-increment is used in the list. Playing around, you could do something like:

print "$_ and $i\n" for (\($i = 1, $i++, $i++));
print \$i , "\n";

__OUTPUT__
SCALAR(0x8176c44) and 3
SCALAR(0x815dcc0) and 3
SCALAR(0x815dad4) and 3
SCALAR(0x8176c44)

This shows that the first element of the list is the $i variable, and that perl generates two temporary scalars to hold the values of $i prior to increment. Also, the entire list is evaluated before iteration.

Interesting stuff you found here!