in reply to Re: Re: Re: For loop problem
in thread For loop problem

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

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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--;

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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

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Re: Re: Re: Re: Re: For loop problem
by chipmunk (Parson) on Jun 04, 2001 at 06:05 UTC
    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.
    Post-increment and -decrement have to squirrel away their value of $x, because they change the value of $x in place but return the previous value.

    Pre-increment and -decrement, on the other hand, change the value of $x in place and then return the new value. Returning a pointer to the location of $x, rather than to a copy of the value in $x, is an optimization. 

    $x = 3;
print ++$x/$x--;

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    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.
    You appear to have the order of evaluation all wrong. The operands to /, this case ++$x and $x--, must both be evaluated before the division. What you describe sounds more like: (++$x/$x)-- except, of course, that it's a syntax error.

    By the way, you'll see the same results with other operations that modify variables in place, such as +=. print $x, $x += 1; Creating a separate copy of every value in every variable, to prevent such effects, would cost a lot in execution time and memory usage. It could also get very complicated with constructions like this: print +(($x+=1)+=1) + ($x+=$x++);

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[d/l]
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      "You appear to have the order of evaluation all wrong."

      Sorry, I should have been more precise. What I was emphasising was that $x-- was still 4 (as it returned the previous value), not that it didn't occur before the division.

      Thanks for pointing out the pointers {g}. I understood the refference passing, I was just getting hung up on left to right evaluation as if it were a stream; not the line completely evaluated, then all the pointer calls.

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