Using a pointer to point to a certain row in c-CodePudding

If I have array a, how would I set a pointer to the first row?

double a[2][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}};

CodePudding user response：

You can declare a pointer to a row and initialize it to point to the first row with the following line:

double (*p_first_row)[4] = &a[0];

Due to array to pointer decay, you can also write:

double (*p_first_row)[4] = a;

The parentheses are necessary, because the declaration

double *p[4];

declares an array of pointers, whereas the declaration

double (*p)[4];

declares a pointer to an array.

CodePudding user response：

Just dereference it normally as you would do to any pointer. *(a 0) gives you the first row of the matrix, and *(a i) will give you the i-th row in the 2D array.

A sample code to get the first element in each row of your 2d array would look like this.

double a[2][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}};

for (int i = 0; i < 2; i   ) {
  printf("%lf ", *(a   i)[0]);
}

Output:

1 5

CodePudding user response：

If you have a multi-dimensional array like for example

T a[N1][N2][N3][N4];

where T is some type specifier and N1, N2, N3, N4 are some positive integers then to make a pointer to the first element of the array just change the left most dimension to asterisk like

T ( *p )[N2][N3][N4] = a;

In this declaration the array designator a is implicitly converted to a pointer to its first element.

If you want to get a pointer to the i-th (0 <= i < N1) element of the array (that is an array of the type T[N2][N3][N4]) you can write

T ( *p )[N2][N3][N4] = a   i;

T ( *p )[N2][N3][N4] = a;
p  = i;

Here is a demonstration program.

#include <stdio.h>

int main( void )
{
    double a[2][4] = 
    {
        {1, 2, 3, 4}, 
        {5, 6, 7, 8}
    };

    for ( double ( *row )[4] = a; row != a   2;   row )
    {
        for ( double *p = *row; p != *row   4;   p )
        {
            printf( "%.1f ", *p );
        }

        putchar( '\n' );
    }
}

The program output is

1.0 2.0 3.0 4.0 
5.0 6.0 7.0 8.0