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cpointersmatrix
shurrok
shurrok

Reputation: 831

How to print matrix in C using pointers?

I cannot figure out why the output of my program is so strange. I just wanted to print matrix with pointers, but what I get is:

1 7 3 8 9
10 8 9 3 4

1 7 3 8 9
7 3 8 9 10

What am I doing wrong here?

#include<stdio.h>
#define NK 5
#define NW 2
int sum(int *w);
int main(void) {

    srand(time(NULL));
    int T[NW][NK];
    int i, j;
    for (i = 0; i<NW; i++) {
        for (j = 0; j<NK; j++) {
            T[i][j] = rand() % 10 + 1;
            printf("%d ", T[i][j]);
        }
        printf("\n");
    }
    int *wsk = T;

    printf("\n");
    sum(wsk);

    return 0;
}
int sum(int *w) {
    int i, j;
    int suma = 0;
    printf("\n");
    for (i = 0; i<NW; i++) {
        for (j = 0; j<NK; j++) {
            printf("%d ", *((w + i)+j));
        }
        printf("\n");
    }

}

Upvotes: 1

Views: 1956

Answers (1)

chqrlie
chqrlie

Reputation: 144770

If the geometry is fixed, you can just declare the argument with the proper type:

int sum(int w[NW][NK]) {
    printf("\n");
    for (int i = 0; i < NW; i++) {
        for (int j = 0; j < NK; j++) {
            printf("%d ", w[i][j]);
        }
        printf("\n");
    }
}

If you insist on passing a pointer to a linearized version:

int sum(int *w) {
    printf("\n");
    for (int i = 0; i < NW; i++) {
        for (int j = 0; j < NK; j++) {
            printf("%d ", w[i * NK + j]);
        }
        printf("\n");
    }
}

Upvotes: 2

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