Trying to pass 2d array to function in C

Question

#include <stdio.h>

void triU(double **U, double *b, int n)
{
    n--;
    b[n] /= U[n][n];
    for(int i = n - 1; i >= 0; i--)
    {
        double aux_sum = 0;
        for(int j = i + 1; j <= n; j++)
            aux_sum += (U[i][j] * b[j]);
        b[i] = (b[i] - aux_sum) / U[i][i];
    }
}

int main()
{
    int n = 3;
    double U[][n] = {{5, -5, 10}, {0, 2, 4}, {0, 0, -1}};
    double b[] = {25, 16, -2};
    triU(U, b, n);
    for(int i = 0; i < n; i++)
        printf("%le \n", b[i]);
    return 0;
}

This is the code for what I am trying to do. I am supposed to make a method with the parameters written in that way to pass the two matrices.

However I keep getting the following error.

triU.c: In function ‘int main()’:
triU.c:21:10: error: cannot convert ‘double (*)[n]’ to ‘double**’
   21 |     triU(U, b, n);
      |          ^
      |          |
      |          double (*)[n]
triU.c:3:20: note:   initializing argument 1 of ‘void triU(double**, double*, int)’
    3 | void triU(double **U, double *b, int n)
      |           ~~~~~~~~~^

I would appreciate any help on how I am supposed to pass arrays by pointers in C. Thank you very much.

score 0 · Answer 1 · edited Nov 05 '22 at 09:50

In this call of the function triU:

triU(U, b, n);

the two-dimensional array U is implicitly converted to pointer to its first element of the type double ( * )[3]. But the corresponding function parameter has the type double ** and there is no implicit conversion between the pointer types. So the compiler issues a message.

If your compiler supports variable length arrays (and your compiler support VLAs) then declare the function like:

void triU( size_t n, double ( *U )[n], double *b );

or:

void triU( size_t n, double U[][n], double *b );

and call it like:

triU( n, U, b );

Otherwise declare the function like:

void triU( double ( *U )[3], double *b, size_t n );

or:

void triU( double U[][3], double *b, size_t n );

and call it like:

triU( U, b, n );

Pay attention to that this for loop:

    for(int j = i + 1; j <= n; j++)
        aux_sum += (U[i][j] * b[j]);

can invoke undefined behavior when j is equal to n because the valid range of indices is [0, n).

Also bear in mind that the length modifier in the format string in this call:

printf("%le \n", b[i]);

is redundant and has no effect. You may write:

printf("%e \n", b[i]);

Trying to pass 2d array to function in C

1 Answers1