how to assign 2 d array to pointer to an array in

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I have a pointer to an array (2d array). I like to assign 2d array to a pointer

  int (*p)[2]={{1,2}};//SegFault
  int (*p)[2]=(int **){{1,2},{3,4}}; //should it be same as above? but also causing segFault/
  int (*p)[2]=(int[][]){{5,4},{5,6}};//should it be same as above two? but also causing segFault

And how can I access the values. should it be correct *(*(p) to access of first array and first value. // Willl it also work p[0][0] I think so, but please clarify that are these access methods are purely how C is designed or Is there any memory matters if I use *(*(p)) or p[0][0]

multidimensional-array pointers initialization implicit-conversion c
2 Answers

This declaration

int (*p)[2]={{1,2}};

is syntactically incorrect. Pointers are scalar objects and they may be initialized by a single expression optionally enclosed in braces.

In this declaration

int (*p)[2]=(int **){{1,2},{3,4}}

you are trying to use a compound literal but again the initialization is incorrect because an object of the type int ** is a scalar object and my be initialized by a single expression.

In this declaration with a compound literal

int (*p)[2]=(int[][]){{5,4},{5,6}};

there is used an incorrect type specifier int[][] for a two-dimensional array because the size of elements of the array is unknown. You could write for example

int (*p)[2]=(int[][2]){{5,4},{5,6}};

That is you need to have an array that you will assign to a pointer. If you do not have such an array you can use a compound literal. For example

int (*p)[2] = (int[2][2]){{1,2},{3,4}};

or

int (*p)[2] = (int[][2]){{1,2},{3,4}};

Here is a demonstrative program.

#include <stdio.h>

int main(void) 
{
    enum { N = 2 };
    
    int ( *p )[N] = ( int[N][N] ){ { 1, 2 }, { 3, 4 } };
    
    for ( size_t i = 0; i < N; i++ )
    {
        for ( size_t j = 0; j < N; j++ )
        {
            printf( "%d ", p[i][j] );
        }
        putchar( '\n' );
    }
    putchar( '\n' );
    
    int a[N][N] = { { 1, 2 }, { 3, 4 } };
    p = a;
    
    for ( size_t i = 0; i < N; i++ )
    {
        for ( size_t j = 0; j < N; j++ )
        {
            printf( "%d ", p[i][j] );
        }
        putchar( '\n' );
    }
    putchar( '\n' );
    
    int b[N] = { 1, 2 };
    p = &b;
    
    for ( size_t i = 0; i < N; i++ )
    {
        printf( "%d ", ( *p )[i] );
    }
    putchar( '\n' );
    
    return 0;
}

The program output is

1 2 
3 4 

1 2 
3 4 

1 2
user_28772410
  • {{1,2}}; is an array initializer, not an array. So the code is invalid C - you can't initialize a pointer like that. It's the wrong type int and the wrong number of initializers.
  • (int **){{1,2},{3,4}}; is a compound literal of type int**. But again, the initialization of a pointer like this is invalid C.
  • (int[][]){{5,4},{5,6}}; is an array of incomplete type. It can't be initialized. So this too is invalid C.

All of your problems originate from ignoring compiler warnings. In case of gcc-like compilers I strongly recommend to change your settings to -std=c11 -pedantic-errors. Then you'll get compiler errors when you write invalid C, instead of just warnings.

To answer the question, the correct way is:

int (*p)[2] = (int[2][2]){ {1,2}, {3,4} };

This creates a local scope 2D array with the same scope as the pointer p. This is equivalent to:

int arr[2][2] = { {1,2}, {3,4} };
int (*p)[2] = arr;
user_5845180
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