for(++i;++i;++i) the second argument is < or <=?

Question

In this for loop statement

#include<stdio.h>

int main()
{
  static int i; 
  for(++i;++i;++i) { 
    printf("%d ",i); 

    if(i==4) 
       break; 
 } 
 return 0; 

}

Variable i is at first 0. The arguments in the for-loop at 1st round are
1st ++i: i = 0 + 1 = 1
2nd ++i: i=1+1=2
So, in first loop I have this for(i=1; i<2; ++i); or for(i=1; i<=2; ++i);?

EDIT I found this example online in a test about C. I run this (inside the for-loop , I have a break point so after some loops it breaks) but I was just guessing the behavior of that so I asked it here to be sure. I am learning now C so stupid questions exists for me. Its better to ask, than not.

You have neither `<` nor `<=` in your code, so why would you expect it to behave like either? — Oliver Charlesworth, Jan 20 '14 at 09:43
@nhahtdh: at some point, `i` will overflow, and `++i` will be `0`, and the loop will break — Elias Van Ootegem, Jan 20 '14 at 09:44
Why do you want to torture yourself with dire code like this? — David Heffernan, Jan 20 '14 at 09:44
You have neither: what you have is (implicitly) `++i != 0`. Since your `i` is incrementing, this will hit undefined behavior when it increments past `INT_MAX`. — user4815162342, Jan 20 '14 at 09:45
Would this even compile? The second condition in a for-loop is of bool type. `++i` IS NOT a bool type. — waka, Jan 20 '14 at 09:49
@Elias or not. Undefined behaviour is undefined. See here for an example: http://stackoverflow.com/questions/7682477/why-does-integer-overflow-on-x86-with-gcc-cause-an-infinite-loop. There is no integer overflow in C. — R. Martinho Fernandes, Jan 20 '14 at 09:50
I run it but I just guess the result!! I wanted to be sure about the behavior of this @OliCharlesworth — yaylitzis, Jan 20 '14 at 09:50
@R.MartinhoFernandes: Fair point... bitten by undefined behaviour once more... — Elias Van Ootegem, Jan 20 '14 at 09:50
@waka, remember that this is C, C does not have a type system — Shoe, Jan 20 '14 at 09:50
When I first used the for-loop I saw examples with `<` and `<=` etc.. I didn't realize that the 2nd argument was just a condition. @OliCharlesworth — yaylitzis, Jan 20 '14 at 09:56

Dipto · Accepted Answer · 2014-01-20T10:06:58.873

In the second argument it is actually ++i!=0, The loop is interpreted as

for(++i;++i!=0;++i)

If you start with a positive i or 0, it will be an infinite loop and will invoke undefined behavior when i reaches INT_MAX.

If i was -Ve initially the loop may stop at a defined run.

EDIT: As you changed your question, Your code will not crash, but you can clearly understand the dry-run by replacing the second ++i with ++i!=0.

So the 1st iteration becomes:

(i=1;2!=0;++i/*this will execute later*/)

2nd iteration becomes:

i=3 //this is from the 1st iteration last part.
(/*initialization is done 1st time only*/;4!=0;++i/*again, this will execute after this iteration*/)

doug65536 · Answer 2 · 2014-01-20T10:16:20.440

It will print 2 4.

Before the for loop, i will be 0. It hasn't been assigned anything yet, and static variables are guaranteed to be zero initialized before they are first used.

It will execute the first ++i in the for loop, since that expression is evaluated once at the beginning of the loop. i will be 1.

It will execute the second ++i, because that is evaluated BEFORE every loop to see if it should run an iteration of the loop. i will be 2.

It will run the loop body. This will print 2.

The if condition won't be true so it won't break.

It will execute the third ++i in the for loop statement, since it evaluates that AFTER every iteration. i will be 3.

It will execute the second ++i again, since it needs to see if it needs to run another loop. It will be nonzero, so it will run another loop. i will be 4.

It will print 4.

The if condition will be true, it will break out of the loop.

However, it is a nonsense way to do it. This is a more appropriate way to do that:

int i;
for (i = 2; i <= 4; i += 2)
    printf("%d ", i);

or better yet:

printf("2 4 ");

I know that this is very stupid way to do this. I just found the example in an online test about C and I was curious about that. — yaylitzis, Jan 20 '14 at 10:12
Hopefully my walk-through helps you understand what those three things are in the `for` statement. — doug65536, Jan 20 '14 at 10:15

score 1 · Answer 3 · answered Jan 20 '14 at 09:57

   static int i;

While the C standard guarantees that variables with static storage duration are initialized to 0, you should not abuse that. Always initialize your variables, either at the line where they are declared or in runtime. So change this to static int i=0;

The first ++i is indeed equivalent to having i=1 there. Esentially your loop does this:

for(i=1; loop_until_program_crash; i++)

If you have a break inside the loop, then the loop is likely poorly written. If you know in advance when the loop should end, then that condition should be inside the for loop condition. If you don't know in advance, then use a while loop instead.

score 0 · Answer 4 · answered Jan 20 '14 at 09:48

0

It should be :

for(i=1; i<=2; ++i);
static int i=0;
for(++i;++i<=2;++i)
{
      printf("4rth :%d\n",i);
}

see : http://ideone.com/TGLYlL

answered Jan 20 '14 at 09:48

shifu

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for(++i;++i;++i) the second argument is < or <=?

4 Answers4