Thread: break in for loop not working

Originally Posted by Sipher

C code is quite hackable, isn't it?

One wrong move and a black hole sprouts before you!

Yep... but I didn't much mind the trip to Alpha Centauri... they have excellent coffee

Code:

matrix[10][10]={0}; //initial declaration of my matrix
int x = 'x';

for(i = 0; i <10; i++){
       for(j = 0; j<cols; j++){
        scanf("%d", &matrix[i][j]);
        if(matrix[i][j] == x)
           break;
                 
             }
     }

it doesn't matter how I do it, the 'i' counter never stops. Even if I moved the if statement after the following bracket, it will just make it worse. When I compile and then run it it gives me extra values (0's in this case). For example I set 'j' =2. Then I enter 4 values, 1 through 4. It would give me 1,2, 3, 4, 0, 0, 0, 0, 0,0

Originally Posted by Sipher

C code is quite hackable, isn't it?

One wrong move and a black hole sprouts before you!

So true, so true...

Code:

int main(void)
{
	const size_t rows = 4, cols = 3;
	int mat[rows][cols] = {{0}};
	int* mb = ( int*)mat, * me = mb + sizeof(mat) / sizeof(int);
	while(mb != me)
		if(scanf("%d", mb++) != 1)
			break;
	for(size_t r = 0; r < rows; ++r)
		for(size_t c = 0; c < cols; ++c)
			printf("mat[%d][%d]: %d\n", r, c, mat[r][c]);		
}

Originally Posted by Sipher

Did you even read our posts so far?

Naaa... he waited for gardhr to post a working solution so he could cut and paste it into his own code...

Originally Posted by CommonTater

Naaa... he waited for gardhr to post a working solution so he could cut and paste it into his own code...

Well, nope. If I were waiting just for the code I would have just asked for it since the beginning. But that's no the point if I won't learn programming >_> newbie

Originally Posted by Sipher

Did you even read our posts so far?

Yes I have, I tried putting the break within just the first for loop and it didn't work. I tried testing for the output/input but I didn't get it right either. I got rid of the if(isalpha()) because like said before, I don't need it. I know that for loops are intended to go through a number of iterations. Just a question to get out of the doubt. Would a while loop be better in this case?

ok, the problem is stopping the 'i' from going all the way to ten. I tried breaks in different places. If I put it within just the 'i' for loop, it won't let me enter enough numbers.
I put a printf statement to print 'i'. If I do the following, this happens

Code:

for(i = 0; i < 10; i++){
       for(j = 0; j<cols; j++){
         scanf("%d", &matrix[i][j]);
            }
         if(matrix[i][j] != 1)
           break;
       }

when run with gcc,

Code:

 
enter the number of columns you want your matrix to be
the number can't exceed 10: 2
you can enter up to 20 items:
please enter the numbers you want in your matrix, enter each number by dividing them with spaces or by pressing enter
enter a non numeric value to stop entering values
1
2

i = 0

***this is your matrix***
(doesn't display anything)

this what happens when I put the break before

Code:

enter the number of columns you want your matrix to be
the number can't exceed 10: 2
you can enter up to 20 items:
please enter the numbers you want in your matrix, enter each number by dividing them with spaces or by pressing enter
enter a non numeric value to stop entering values
1
2
3
4
x

i = 10
***this is your matrix***
  1  2
  3  0
  4  0
  0  0
  0  0

@torquemada...

Did you even try the suggestion of making it into a function and using return to exit both loops?

Originally Posted by GReaper

Just before the break set the i to 10 (or greater).
Then the first for loop will stop.

Dylan

I tried it but the problem is that it won't let me enter numbers pass 2. It might not be the best implementation.

Code:

   for(i = 0; i < 10; i++){
       for(j = 0; j<cols; j++){
         scanf("%d", &matrix[i][j]);
         if(matrix[i][j] != 1)
          i = 10; 
        }
      
      if(i >= 10)
        break;
      }

output

Code:

enter the number of columns you want your matrix to be
the number can't exceed 10: 2
you can enter up to 20 items:
please enter the numbers you want in your matrix, enter each number by dividing them with spaces or by pressing enter
enter a non numeric value to stop entering values
1
2
(it won't let me enter more numbers)

***this is your matrix***
  1  2
  0  0
  0  0
  0  0
  0  0

Sorry for the late reply.
I just wanted to say thanks to every one for their time

I figured something out
Instead of letting 'i' be my counter I created another variable "counter" to count 'i' but it wouldn't go higher once the break is reached in the loop. So I stored that value and used it as a counter. It might not sound like the best idea to you but it works for me.
Thanks again