Trying to pass 2d array to function in C [duplicate] - c

Question

This question already has answers here:
How to pass a 2D array by pointer in C?
(5 answers)
Closed 4 months ago.
#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.

Answer 1

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]);