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Re: Re: Mysterious for behavior

by Roy Johnson (Monsignor)
on Apr 21, 2004 at 19:12 UTC ( #347101=note: print w/replies, xml ) Need Help??

in reply to Re: Mysterious for behavior
in thread Mysterious for behavior


Update 2: Because of the explanations given, I understand the phenomenon, but I think it can be explained more explicitly than has been done, so I'm going to give it a shot:

First of all, all the expressions will be evaluated when the list is built. That is, before any iteration happens, all of those assignments have been executed.

The next thing to understand is that lists may include ordinary values and lvalues; an assignment yields an lvalue, though an expression of which an assignment is just part (e.g., 0+($x=3)) is an ordinary value. What is stored in an lvalue can change even after the lvalue is inserted into the list, so its value depends on when you look at it.

The aliasing of a for loop is not fundamental to this. Simply assigning the same list to an array demonstrates the same phenomenon:

my $x; my @a = ($x=1, $x+=1, $x-=5); print "@a\n"; #yields: #-3 -3 -3
One final interesting note: postincrement does not yield an lvalue, but preincrement does.

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Re: Re: Re: Mysterious for behavior
by Belgarion (Chaplain) on Apr 22, 2004 at 22:05 UTC

    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!

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