Pointer Arithmetics and Arrays

[GokhanK]

I am trying to figure out how pointer arithmetics work on arrays. However I couldn't make an example work.

Firstly I start with a one dimensional array and the example works;

#include <stdio.h>

int main(void){
    int array[2];
    int *ptr=array;
    *(ptr+1)=1000;
    printf("%d",array[1]);
    return 0;
}

After that I add one more dimension and it doesn't show the result I want;

#include <stdio.h>

int main(void){
    int array[2][2];
    int (*ptr)[2]=array;
    *(ptr+1)=999;
    printf("%d",array[1][0]);
    return 0;
}

Why doesn't this increase the address by one *ptr[2]? Isn't it possible to perform pointer arithmetics with pointer to arrays?

[jimblumberg]

Does that second snippet even compile? I get the following error when I try: main.c|6|error: assignment to expression with array type|


Jim

[GokhanK]

My compiler doesn't give me an error and shows a console window but the one below may compile well. However it still doesn't work.

#include <stdio.h>
 
int main(void){
    int array[2][2];
    int (*ptr)[2];
    ptr=array;
    *(ptr+1)=999;
    printf("%d",array[1][0]);
    return 0;
}

[MutantJohn]

I think you need one more dereference.

ptr[1][0] = 999;

[GokhanK]

I think you need one more dereference.

ptr[1][0] = 999;

Thanks for the hint, one more dereference solves the problem. I will think about the logic.

#include <stdio.h>
  
int main(void){
    int array[2][2];
    int (*ptr)[2];
    ptr=array;
    *(*(ptr+1))=999;
    printf("%d",array[1][0]);
    return 0;
}