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### Re: Re: Re: For loop problem

by japhy (Canon)
 on Jun 03, 2001 at 23:19 UTC ( #85357=note: print w/replies, xml ) Need Help??

in reply to Re: Re: For loop problem

The problem is that the list (\$a .. ++\$a) is not evaluating a list like you think it is. It is not doing:
```\$start = \$a;
\$end = ++\$a;
for (\$_ = \$start; \$_ <= \$end; \$_++) { ... }
But rather, it is evaluating its endpoints, and then constructing the list from that. That is the cause of your problem:
```for (++\$x .. \$x++ . \$x) { ... }
# starting at ++\$x
# ending at \$x++ . \$x
# THE END HAS MODIFIED THE START
You can witness a similar result from:
```print ++\$x, ++\$x, ++\$x;  # 333
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.

Knowing this, the output of this program makes sense to me:

```\$x = 3;
print ++\$x/\$x--;  # 3/4 => 0.75
Why is that? Well, the numerator is ++\$x, and the denominator is \$x--; but in subtracting 1 from \$x in the denominator, we have now altered the numerator back to its original state!

I've worked out a solution to your 1 .. 12 loop, then.

```for (++\$x .. \$x++ . \$x--) { ... }
# or
for (++\$....\$.++.\$.--) { ... }
See if you can work out why the starting point is 1.

japhy -- Perl and Regex Hacker

Replies are listed 'Best First'.
Re: Re: Re: Re: For loop problem
by Arguile (Hermit) on Jun 04, 2001 at 00:26 UTC
```print ++\$x, ++\$x, ++\$x; #333
print ++\$x, \$x++, ++\$x; #313

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"

```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

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--;

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
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--;
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++);

"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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