why does this for loop iterating over 2d array not work

Question

#include <stdio.h>
#include <string.h>
#include <stdlib.h>


int main(void)
{

    int a[4][5] =
    {
       {1,1,0,1,0},
       {0,0,0,0,0},
       {0,1,0,0,0},
       {1,0,1,1,0}
    };

    int *ptr = &a[0][0];

    for (; ptr < ptr+19; ptr++)
    {
        printf("%d ", *ptr);
    } 
  
    return 0;
    
}

Why does this code not work? as far as i know 2D arrays are stored in row major order so like this:

[1,1,0,1,0]-[0,0,0,0,0]-[0,1,0,0,0]-[1,0,1,1,0]

So if i started at &a[0][0] and incremented it by 1 each time until it has been incremented to the last element why does this not work?

ptr is always less than ptr+19, at least until the pointer runs off the end of the address space and wraps around, but it is likely to be pointing to invalid memory before that happens. — Ian Abbott
– Ian Abbott, Commented Jul 7, 2021 at 11:35
"Does not work" is not a useful statement. What does it do? — Andrew Henle
– Andrew Henle, Commented Jul 7, 2021 at 11:35
Apart from the obvious bug ptr < ptr+19 you aren't allowed to de-reference arrays in this manner. You go out of bounds in the array access and it's undefined behavior. The compiler is allowed to go nuts and generate strange code or a crashing program. — Lundin
– Lundin, Commented Jul 7, 2021 at 14:26

MikeCAT · Accepted Answer · 2021-07-07 11:36:11Z

2

Aside from the danger caused from accessing data with pointers for different type, the loop condition is wrong.

ptr+19 moves with ptr and the condition ptr < ptr+19 will always be true (until overflow occurs).

Your code will be better if you retain the origin pointer and use that for stop condition.

    int *ptr = &a[0][0];
    int *start_ptr = ptr;

    for (; ptr < start_ptr+19; ptr++)
    {
        printf("%d ", *ptr);
    }

answered Jul 7, 2021 at 11:36

MikeCAT

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9 Comments

programmer29874 Over a year ago

wait a minute the ptr+19 adds 19 to the pointer each time?!

MikeCAT Over a year ago

@programmer29874 Yes... or maybe no because the condition ptr < ptr+19 may be turned into an unconditional "true" by the optimizer.

programmer29874 Over a year ago

wow i did not realize that Mike thanks a lot, you seem to be pro at C could you recommend me something to get better at C? maybe a book or some materials ;P

Ian Abbott Over a year ago

@programmer29874 Try it with int: for (int x = 0; x < x + 19; x++) printf("%d ", x);. How many iterations do you think it should do?

datenwolf Over a year ago

@programmer29874: this is not really a C issue. It's just that the mathematical expression x < x + y, where 0 < y always holds true, regardless of if it's part of a program or just some manually executed arithmetic. The conditional part of a loop is evaluated whole on each iteration. The compiler is not a magical beast that sees an expression like a < b and thinks "oh, I'll have to reevaluate b and only b each time".

|

Edgen · Accepted Answer · 2021-07-07 12:23:17Z

0

Alternative approach

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(void)
{

    int a[4][5] =
    {
       {1,1,0,1,0},
       {0,0,0,0,0},
       {0,1,0,0,0},
       {1,0,1,1,0}
    };

    int len = sizeof(a)/sizeof(a[0][0]) ;
    int * ptr = &a[0][0] ;
    while(len--)
    {
        printf("%d ", *ptr++);
    }
  
    return 0;
    
}

answered Jul 7, 2021 at 12:23

Edgen

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Comments

Vlad from Moscow · Accepted Answer · 2021-07-07 12:39:13Z

0

In this for loop

for (; ptr < ptr+19; ptr++)

the value of the expression ptr is always less than the value of the expression ptr + 19 (provided that there is no overflow in the last expression). Moreover the pointer ptr is incremented after each iteration of the loop.

So using this loop you will get access to memory outside the array a that results in undefined behavior.

What you need is the following for statement

for ( ; ptr < ( int * )a + sizeof( a ) / sizeof( **a ); ptr++)

Here is a demonstrative program.

#include <stdio.h>

int main(void) 
{
    int a[4][5] =
    {
       {1,1,0,1,0},
       {0,0,0,0,0},
       {0,1,0,0,0},
       {1,0,1,1,0}
    };

    for ( int *ptr = ( int * )a; ptr < ( int * )a + sizeof( a ) / sizeof( **a ); ptr++ )
    {
        printf( "%d ", *ptr );
    }
    
    putchar( '\n' );
    
    return 0;
}

The program output is

1 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 1 1 0

edited Jul 7, 2021 at 12:39

answered Jul 7, 2021 at 11:53

Vlad from Moscow

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2 Comments

programmer29874 Over a year ago

i have one question does the condition ptr < ptr+19 add 19 to ptr each time its evaluated?

Vlad from Moscow Over a year ago

@programmer29874 ptr is incremented in the third expression of the for loop ptr++. In the second expression (condition) of the for loop there are compared two expressions ptr and ptr + 19.

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