How can I copy a struct member array to a variable?

Question

I have the following basic C example:

#include <stdio.h>

struct Example {
    double arr[4][4];
    int size;
};

int main()
{
    struct Example e = {
        {{1., 0., 0., 0.},
         {0., 1., 0., 0.},
         {0., 0., 1., 0.},
         {0., 0., 0., 1.}},
         4
    };
    double** arr = e.arr;

    printf("%8.5f %8.5f %8.5f %8.5f\n", arr[0][0], arr[0][1], arr[0][2], arr[0][3]);
    printf("%8.5f %8.5f %8.5f %8.5f\n", arr[1][0], arr[1][1], arr[1][2], arr[1][3]);
    printf("%8.5f %8.5f %8.5f %8.5f\n", arr[2][0], arr[2][1], arr[2][2], arr[2][3]);
    printf("%8.5f %8.5f %8.5f %8.5f\n", arr[3][0], arr[3][1], arr[3][2], arr[3][3]);
}

What do I have to do to make the printf successfully print out the values in the matrix? If I declare arr as double** I get a segfault. If I try double* then it complains when I try to do double indexing. I've also tried double arr[4][4] = e.arr, but the compiler just tells me that it's an invalid initializer. What's the proper way to do this?

(I realize size is redundant in this example, I just wanted the struct to have more than one member.)

You can use a pointer to array like this double (*arr)[4] = e.arr. Note that e.arr is the same as &e.arr[0], but &e.arr[0] is the address of an entire array with 4 doubles (a line of the 2D matrix), so you need a pointer that can point to an entire array of length 4. — isrnick
– isrnick, Commented Jun 5, 2020 at 4:18
There is no copy of the array being made here, you are actually storing the address of the array in a pointer variable, so the pointer points to the original array, and the values printed belong to the original array, not a copy of it. — isrnick
– isrnick, Commented Jun 5, 2020 at 4:49
c-faq.com/aryptr will be helpful background information for you here. — Karl Knechtel
– Karl Knechtel, Commented Jun 5, 2020 at 5:08

AndersK · Accepted Answer · 2020-06-05 05:03:02Z

1

You can do like this since the values of double arr[4][4] are consecutive in memory

double* arr = e.arr[0]; // points to first value

printf("%8.5f %8.5f %8.5f %8.5f\n", arr[0], arr[1], arr[2], arr[3]);
printf("%8.5f %8.5f %8.5f %8.5f\n", arr[4], arr[5], arr[6], arr[7]);
printf("%8.5f %8.5f %8.5f %8.5f\n", arr[8], arr[9], arr[10], arr[11]);
printf("%8.5f %8.5f %8.5f %8.5f\n", arr[12], arr[13], arr[14], arr[15]);

or alternatively, as @ismick pointed out in his comment you can write it like

double (*arr2)[4] = e.arr ;

printf("%8.5f %8.5f %8.5f %8.5f\n", arr2[0][0], arr2[0][1], arr2[0][2], arr2[0][3]);
printf("%8.5f %8.5f %8.5f %8.5f\n", arr2[1][0], arr2[1][1], arr2[1][2], arr2[1][3]);
printf("%8.5f %8.5f %8.5f %8.5f\n", arr2[2][0], arr2[2][1], arr2[2][2], arr2[2][3]);
printf("%8.5f %8.5f %8.5f %8.5f\n", arr2[3][0], arr2[3][1], arr2[3][2], arr2[3][3]);

if you want that syntax.

edited Jun 5, 2020 at 5:03

answered Jun 5, 2020 at 4:54

AndersK

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Comments

Shubham · Accepted Answer · 2020-06-05 05:59:47Z

0

Alternative to the answer given by @AndersK, you can also try this.

#include <stdio.h>

struct Example {
    double arr[4][4];
    int size;
};

int main()
{
    int i, j;
    struct Example e = {
        {{1., 0., 0., 0.},
         {0., 1., 0., 0.},
         {0., 0., 1., 0.},
         {0., 0., 0., 1.}},
         4
    };
    double *arr = (double *)e.arr;

    for(i = 0; i < 4; ++i)
    {
        for(j = 0; j < 4; ++j)
            printf("%8.5f ", *(arr+i*4+j));
        putchar('\n');
    }
    return 0;
}

In this we are pointing to the first block of memory where e.arr[0] is stored, then we are traversing through the consecutive memory location using this expression (arr+i*4+j).

answered Jun 5, 2020 at 5:59

Shubham

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