for loop

[Unregistered]

Hi
Below is a small piece of code posted to the board a few days ago, can some one explain to me what is happening to it??
I realise that it is increaseing by three but what makes 'i' print three idetical numbers but 'j' prints numbers sequentialy.
Are the two for's not the same??

#include <stdio.h>

int main(void)
{

int i, j ;

for (i = 1; i <= 3; i++)
for (j = 1; j <= 3; j++)
printf("%d ------ %d\n", i, j);
system("pause");
return 0;
}

the way the numbers are displayed seems to be controled by the 'i' for !!. If I alter the numbers in the for's 'i' I get
two diffrent displays but governed by what is in the 'i' for.

#include <stdio.h>

int main(void)
{

int i, j ;

for (i = 1; i <= 3; i++)
for (j = 1; j <= 6; j++)
printf("%d ------ %d\n", i, j);
system("pause");
return 0;
}

the code above gives :-
1 1
1 2
1 3
1 4
1 5
1 6
2 1
2 2ect,ect up to a count of three on the left. But six digits on the right.why does it not display six out puts on the right.If I put six in the 'i' for it displays six units(left and right)

[-KEN-]

Ok, you obviously need a bit of help with loops...

What's happening is that first, your first loop goes into action, which activates the second loop, which is set to increment 6 times (so you get 6 values) then it returns to the topmost loop, which repeats the process 2 more times (it's set to do it 3 times, so you get 3 values) you get the value of 1 six times because the top loop won't spin (increment, whatever) again until the bottom loop is finished, which isn't until it loops 6 times. Get it?

[Unregistered]

Hi
Right, did not realise the second 'for' was the statment to the first 'for'. But......................what tells printf to pad out the first 'for'
why do you not get the following display
1 1
2
3
4
5
6
2 1
2
3
4
5
6
3 1
2
3
4
5
6
please fore give my ignorance
I must be doing some thing wrong as the board will not display the numbers right, I hope you under stand what I mean

[Prelude]

If you indent and add blocks it makes a little more sense.

#include <stdio.h> 

int main(void) 
{
  int i, j ; 

  for (i = 1; i <= 3; i++) //exterior loop
  { 
    for (j = 1; j <= 3; j++) //interior loop
    {
      printf("%d ------ %d\n", i, j);  //first print is 1------1
    }
  }
  return 0; 
}

What is happening is that when you finally call your print function, you are already inside the interior loop, this means that the value for i will not change until you meet the condition to exit the interior loop (ie. when j == 3). When you exit the interior loop, the exterior loop will increment by one and then you enter the interior loop again. The same thing happens.

So i will stay the same while j is being incremented, and will only be incremented itself when j is done with it's loop. So your output is:

1--------1
1--------2
1--------3
2--------1
2--------2
2--------3
3--------1
3--------2
3--------3
//done

-Prelude

[bigtamscot]

A classic example of "nested loops".

[Sebastiani]

If you wanted to run the loops independantly, you must do:

for(i=0;i<3;i++)
 { 
  printf("%i \n",i);  
 }
for(j=0;j<3;j++)
 {
  printf("%i \n",j);
 } 

Otherwise, (as in the code you posted) the compiler thinks you meant:

for(i=0;i<3;i++)
 { 
   for(j=0;j<3;j++)
    {
     printf("%i  %i\n",i,j);
    }
 } 

So, no, they cannot share the same printf() statement if you want them to run separately...