I want to pass two matrices as argument. These matrices have different size and i don't understand how i have to do this work:
#include <stdio.h>
#include <stdlib.h>
void f(int m[3][], int n);
int main()
{
int A[3][3]={{1,2,3},{4,5, 6},{7,8,9}};
int B[3][2]={{1,2},{3, 4}, {5, 6}};
f(A, 3);
f(B, 2);
return 0;
}
void f(int m[3][], int n)
{
int i,j;
for(i=0;i<3;i++)
{
for(j=0;j<n;j++)
printf("%5d", m[i][j]);
}
return;
}
How can I do this?
1) In C, if we pass an array as an argument to a function, what actually get passed? The correct option is (b). Explanation: In C language when we pass an array as a function argument, then the Base address of the array will be passed.
We cannot pass the function as an argument to another function. But we can pass the reference of a function as a parameter by using a function pointer. This process is known as call by reference as the function parameter is passed as a pointer that holds the address of arguments.
To pass an entire array to a function, only the name of the array is passed as an argument. result = calculateSum(num); However, notice the use of [] in the function definition. This informs the compiler that you are passing a one-dimensional array to the function.
The only safe way that I know of to do this is to include the matrix dimensions in the parameters, or make some kind of matrix struct
Option A) dimensions as parameters
void f(int **m, int w, int h )
{
int i,j;
for(i=0;i<w;i++)
{
for(j=0;j<h;j++)
printf("%5d", m[i][j]);
}
return;
}
Option B) Use a struct
typedef struct Matrix
{
int w, h;
int** m;
} Matrix;
void f ( Matrix *m )
{
for ( int i = 0; i < m->w; ++i )
{
for ( int j = 0; j < m->h; ++j )
{
printf(%5d", m->m[i][j]);
}
}
}
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