C pointer with 2D array

Question

I'm learning C language, and I'm confused in 2D array pointer.

I have the following declaration

int a[3][5];
int *b[3][3];
int (*c)[3][5];
int *(d[3][5]);
int (*e[3])[5];

could anyone help me clarify

1. if there're valid declaration or not
2. the sizeof a,b,c,d,e (assume on 64-bit machine, address id 8 bytes)
3. what they point to
4. how to access the element in the 2D array

I'm totally confused about the usage of pointer in 2D array.....and I guess some of them are equivalent...but some might not be good practice

Answer 1

These

int a[3][5];
int *b[3][5];

(I think you mean indeed int *b[3][5] instead of int *b[3][3] as it is written in your question) two declarations of two-dimensional arrays.

Elements of the first array have type int. Elements of the second array have type int *.

To access elements of the arrays you can use for example subscripting like

a[i][j] or b[i][j] where i is an index in the range [0,3) and j is an index in the range [0, 5).

For the second array to access objects pointed to by the elements of the array you can use expressions like *b[i][j]

sizeof( a ) is equal to 3 * sizeof( int[5] ) that in turn is equal to 3 * 5 * sizeof( int ).

sizeof( b ) is equal to 3 * sizeof( int *[5] ) that in turn is equal to 3 * 5 * sizeof( int *).

This

int (*c)[3][5];

is a declaration of a pointer to a two-dimensional array of the type int[3][5]. You can write for example

int (*c)[3][5] = &a;

where a is the two-dimensional array declared above.

To access elements of the pointed array you can use this syntax

( *c )[i][j]

This

int *(d[3][5]);

a declaration of a two-dimensional array elements of which have type int *. This declaration is equivalent to the declaration shown above that is to

int *b[3][5];

You may enclose declarators in parentheses. So you could even write the declaration of the array d like

int * (((d)[3])[5]);

This

int (*e[3])[5];

is a declaration of an array with 3 elements of pointers to arrays of the type int[5]. Using the typedef

typedef int T[5]; 

the array declaration can be rewritten like

T * e[3];

Here a demonstrative program that shows how elements of the array e can be accessed.

#include <stdio.h>

int main( void )
{
    int a1[5] = { 1, 2, 3, 4, 5 };
    int a2[5] = { 10, 20, 30, 40, 50 };
    int a3[5] = { 100, 200, 300, 400, 500 };

    int(*e[3])[5] = { &a1, &a2, &a3 };

    for (size_t i = 0; i < sizeof( e ) / sizeof( *e ); i++)
    {
        for (int j = 0; j < sizeof(*e[i]) / sizeof(**e[i]); j++)
        {
            printf( "%3d ", ( *e[i] )[j] );
        }
        putchar( '\n' );
    }

    return 0;
}

The program output is

  1   2   3   4   5
 10  20  30  40  50
100 200 300 400 500