print ++$x, ++$x, ++$x; #333
print ++$x, $x++, ++$x; #313

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Given a left to right evaluation, to me this should produce 123 and 113 respectively.

"You see, the arguments to a function aren't copies of the variables, but rather aliases -- by modifying ONE of the $x's, you've modified the others as well"

I can accept that but what about this: 
my $x=3;
print $x, $x=5, ++$x, $x--, $x, $x++, $x=0;   #0006050
print $x++, $x=5, ++$x, $x--, $x, $x++, $x=0; #3006050

[download]

Again, 3566550 seems the reasonable output (to me) for both accounts. Instead the post-(in|de)crement operators seem to be squireling away their value of $x when it came by them in a left to right evaluation of the terms.

I can see the pattern and work with it, my question is why is it that way? What are the advantages to this type of evaluation? I don't know enough Perl to extrapolate the reasoning behind them.

As one final example consider: 
$x = 3;
print ++$x/$x--;

[download]

Coming into the function x is 3, it gets pre-incremented (now 4), it's divided by x (still 4), x is post-decremented (to 3). Print outputs 1 (4/4) and x leaving the function is 3. This is how it _should_ happen to my mind, instead it's 3/4 as stated earlier.

Maybe it's just my logic that's skewed or I'm showing a glaring flaw in my understanding of how operations need to work internally, but it seems rather convoluted for no good reason.

-- I seek enlightenment

In reply to Re: Re: Re: Re: For loop problem by Arguile
in thread For loop problem by srawls

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