Comments on: can you increment i?

By: Rahul

Rahul — Fri, 05 Oct 2007 20:07:25 +0000

Hi! I tried the code on C and C++ both, and its same in both cases.
int main()
{
int value,i=1;
value=i + i++ + ++i;
printf(”%d”,value);
i=1;
i += i++ + ++i;
printf(” %d”,i);
return 0;
}

Next i tried it with both volatile and without volatile keyword on “i”. Now the catch is:-
1) Without volatile, i.e. with optimization, the answer is 4 7
2) With Volatile, i.e. without optimization, the answer is 4 5

**Will anybody make it clear why the answer in first part is 7 and in other is 5, what effect does optimization have on this???

By: melz

melz — Wed, 29 Jan 2003 22:12:04 +0000

Oddly, I guess 6.

By: blog.tukker.org

blog.tukker.org — Mon, 20 Jan 2003 12:26:49 +0000

[...] p;#8220;i” $i = 1; $i += ++$i + $i++; print $i; For the answer look at the Original post Comments (0) | TrackBack (0) | PingBack (0) Time for this page: 0.054 [...]

By: Qingy

Qingy — Fri, 03 Jan 2003 16:37:23 +0000

I guessed 6 :(
i starts as 1, then it’s added to it’s increment thus 2, but i++; is not counted in the addtion until AFTER the statement, so I got 1 += 2 + 2, thus 5, then the i++ kicks in, giving 6 =\

By: Renee Teunissen

Renee Teunissen — Mon, 09 Dec 2002 22:29:12 +0000

Got 5, as c-code should have given..

Renee.

By: Cessy

Cessy — Sat, 07 Dec 2002 01:25:08 +0000

Whoa…I’m having a headache. Keep those in the source. >:O

By: Rydain

Rydain — Thu, 28 Nov 2002 07:42:43 +0000

I quickly guessed 5 and tested it out in C++ (using a plain old int and compiling with g++). The answer turned out to be 7. If I think about it some more, it will probably make sense.

/me squirts some WD-40 onto the folds of her brain

By: Max

Max — Fri, 22 Nov 2002 16:53:06 +0000

Ah! Oh!
And I didn’t guessed anything! %-(

By: Ling

Ling — Mon, 18 Nov 2002 23:50:35 +0000

i guessed 7 :)

By: Christina

Christina — Sat, 16 Nov 2002 17:49:31 +0000

I guessed 6. ;)

By: The Through

The Through — Wed, 13 Nov 2002 14:58:38 +0000

It depends on C code optimisations too ;o)

if you compile:
main() {
volatile int i;
i = 1;
i += ++i + i++;
printf("result: %i\n", i);
}

you'll see it returns: 3
and if you consider

register int i;

it gives: 7

By: totalementcretin

totalementcretin — Mon, 11 Nov 2002 23:03:34 +0000

garoo: the problem lies with the pre-incrementation stuff, not the post.

By: garoo

garoo — Mon, 11 Nov 2002 17:12:25 +0000

That's the trap, Michel! :))

Try this:
$i = 1; print ($i + $i++);
$i = 1; print ($i++ + $i);

If you were right, both lines would give the same result, because the ++ would be interpreted after the print.
Yet they don't, because the ++ is interpreted right after the $i++ is evaluated.

By: RoyalTS - without cheese

RoyalTS - without cheese — Sun, 10 Nov 2002 22:02:46 +0000

i++, ++i, +=i
can you increment i?

By: michel v

michel v — Sun, 10 Nov 2002 17:28:44 +0000

That’s the trap, Garoo! :)
In fact, 7 is not the result that comes right from the addition. For example if you do “print ($i + ++$i + $i++)”, it will display 6, and then after that statement $i will equal 7.

By: garoo

garoo — Sun, 10 Nov 2002 17:24:55 +0000

To me, that's the way it works:

$i += ++$i + $i++
[4] [1] [3] [2]

[1] i is incremented => $i == 2
[1b] i is evaluated for the addition => $i += 2 + $i++
[2] i is evaluated for the addition => $i += 2 + 2
[2b] i is incremented right after being evaluated => $i == 3
[3] 2 + 2 = 4, so now we have $i += 4, and $i == 3
[4] 3 + 4 = 7

Ok, it's not very clearly explained, but I'm late for an appointment, sorry :)

By: michel v

michel v — Sun, 10 Nov 2002 16:47:09 +0000

Don’t be confused, I may be totally wrong. It all comes down to this question: is the equal sign a sequence stop in PHP? It is not in C, AFAIK.

By: skamp

skamp — Sun, 10 Nov 2002 16:45:46 +0000

OK, now I’m confused :-/

By: skamp

skamp — Sun, 10 Nov 2002 16:43:42 +0000

OK, a quick reading of PHP’s documentation got me the answer concerning the output in that language.
In PHP, the expression is evaluated in the following order :
++$i -> $i = 2
$i++ -> $i = 3
$i + $i -> 3 + 3
$i += -> 1 + 3 + 3

http://docs.skamp.net/php/language.operators.html
http://docs.skamp.net/php/language.operators.increment.html

By: michel v

michel v — Sun, 10 Nov 2002 16:41:15 +0000

This is how I see it operated (at least in PHP):

$i += ++$i + $i++;
[2] [1] [3]

2 += 2 + 2
and then $i++, which is executed after the mathematical statement, makes that 6+1 = 7 :)

Your explanation is maybe how Perl works with that line.