在 C 中使用指针乘以二次矩阵

Question

I have a task where I'm supposed to multiply two quadratic matrices of size n in C, using pointers as function parameters and return value. This is the given function head: int** multiply(int** a, int** b, int n) . Normally, I would use three arrays (the two matrices and the result) as parameters, but since I had to do it this way, this is what I came up with:

#include <stdio.h>
#include <stdlib.h>

int** multiply(int** a, int** b, int n) {
  int **c = malloc(sizeof(int) * n * n);
  // Rows of c
  for (int i = 0; i < n; i++) {
    // Columns of c
    for (int j = 0; j < n; j++) {
      // c[i][j] = Row of a * Column of b
      for (int k = 0; i < n; k++) {
        *(*(c + i) + j) += *(*(a + i) + k) * *(*(b + k) + j);
      }
    }
  }
  return c;
}

int main() {
  int **a = malloc(sizeof(int) * 2 * 2);
  int **b = malloc(sizeof(int) * 2 * 2);

  for (int i = 0; i < 2; i++) {
    for (int j = 0; i < 2; j++) {
      *(*(a + i) + j) = i - j;
      *(*(b + i) + j) = j - i;
    }
  }
  
  int **c = multiply(a, b, 2);

  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      printf("c[%d][%d] = %d\n", i, j, c[i][j]);
    }
  }

  free(a);
  free(b);
  free(c);

  return 0;
}

I have not worked much with pointers before, and am generally new to C, so I have no idea why this doesn't work or what I'd have to do instead. The error I'm getting when trying to run this program is segmentation fault (core dumped) . I don't even know exactly what that means... :(

Can someone please help me out?

Answer 1

There's lots of fundamental problems in the code. Most notably, int** is not a 2D array and cannot point at one.

• i<2 typo in the for(int j... loop.
• i < n in the for(int k... loop.
• To allocate a 2D array you must do: int (*a)[2] = malloc(sizeof(int) * 2 * 2); . Or if you will malloc( sizeof(int[2][2]) ) , same thing.
• To access a 2D array you do a[i][j] .
• To pass a 2D array to a function you do void func (int n, int arr[n][n]);
• Returning a 2D array from a function is trickier, easiest for now is just to use void* and get that working.
• malloc doesn't initialize the allocated memory. If you want to do += on c you should use calloc instead, to set everything to zero.
• Don't write an unreadable mess like *(*(c + i) + j) . Write c[i][j] .

I fixed these problems and got something that runs. You check if the algorithm is correct from there.

#include <stdio.h>
#include <stdlib.h>

void* multiply(int n, int a[n][n], int b[n][n]) {
  int (*c)[n] = calloc(1, sizeof(int[n][n]));

  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        c[i][j] += a[i][k] * b[k][j];
      }
    }
  }
  return c;
}

int main() {
  int (*a)[2] = malloc(sizeof(int[2][2]));
  int (*b)[2] = malloc(sizeof(int[2][2]));

  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      a[i][j] = i - j;
      b[i][j] = j - i;
    }
  }
  
  int (*c)[2] = multiply(2, a, b);

  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      printf("c[%d][%d] = %d\n", i, j, c[i][j]);
    }
  }

  free(a);
  free(b);
  free(c);

  return 0;
}

Answer 2

You need to fix multiple errors here:

1/ line 5/24/28: int **c = malloc(sizeof(int*) * n )

2/ line 15: k<n

3/ Remark: use a[i][j] instead of *(*(a+i)+j)

4/ line 34: j<2

5/ check how to create a 2d matrix using pointers.

#include <stdio.h>
#include <stdlib.h>
  
int** multiply(int** a, int** b, int n) {
  int **c = malloc(sizeof(int*) * n );
  for (int i=0;i<n;i++){
    c[i]=malloc(sizeof(int) * n );
  }
  
  // Rows of c
  for (int i = 0; i < n; i++) {
    // Columns of c
    for (int j = 0; j < n; j++) {
      // c[i][j] = Row of a * Column of b
      for (int k = 0; k < n; k++) {
    c[i][j] += a[i][k] * b[k][j];
      }
    }
  }
  return c;
}

int main() {
  int **a = malloc(sizeof(int*) * 2);
  for (int i=0;i<2;i++){
    a[i]=malloc(sizeof(int)*2);
  }
  int **b = malloc(sizeof(int) * 2);
  for (int i=0;i<2;i++){
    b[i]=malloc(sizeof(int)*2);
  }

  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      a[i][j] = i - j;
      b[i][j] = i - j;
    }
  }

  int **c = multiply(a, b, 2); 

  for (int i = 0; i < 2; i++) { 
    for (int j = 0; j < 2; j++) { 
      printf("c[%d][%d] = %d\n", i, j, c[i][j]); 
    } 
  } 
  free(a);
  free(b);
  free(c);
  return 0;
}

Answer 3

From the updated requirement, the actual function prototype is int *multiply(int *a, int *b, int n); so the code should use a "flattened" matrix representation consisting of a 1-D array of length n * n .

Using a flattened representation, element ( i , j ) of the n * n matrix m is accessed as m[i * n + j] or equivalently using the unary * operator as *(m + i * n + j) . (I think the array indexing operators are more readable.)

First, let us fix some errors in the for loop variables. In multiply :

      for (int k = 0; i < n; k++) {

should be:

      for (int k = 0; k < n; k++) {

In main :

    for (int j = 0; i < 2; j++) {

should be:

    for (int j = 0; j < 2; j++) {

The original code has a loop that sums the terms for each element of the resulting matrix c , but is missing the initialization of the element to 0 before the summation.

Corrected code, using the updated prototype with flattened matrix representation:

#include <stdio.h>
#include <stdlib.h>

int* multiply(int* a, int* b, int n) {
  int *c = malloc(sizeof(int) * n * n);
  // Rows of c
  for (int i = 0; i < n; i++) {
    // Columns of c
    for (int j = 0; j < n; j++) {
      // c[i][j] = Row of a * Column of b
      c[i * n + j] = 0;
      for (int k = 0; k < n; k++) {
        c[i * n + j] += a[i * n + k] * b[k * n + j];
      }
    }
  }
  return c;
}

int main() {
  int *a = malloc(sizeof(int) * 2 * 2);
  int *b = malloc(sizeof(int) * 2 * 2);

  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      a[i * 2 + j] = i - j;
      b[i * 2 + j] = j - i;
    }
  }
  
  int *c = multiply(a, b, 2);

  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      printf("c[%d][%d] = %d\n", i, j, c[i * 2 + j]);
    }
  }

  free(a);
  free(b);
  free(c);

  return 0;
}