2D数组的乘法，使用指向指针的指针 [英] Multiplication of 2D arrays, using pointers to pointers

问题描述

我一直在尝试用C语言实现一个程序，其中我有两个2D数组(使用指针创建)，并且我必须将它们相乘(如相乘矩阵)并将结果存储到第三个数组中.阵列的尺寸由用户指定.我已经完成了代码，但是结果却是错误的.我的配方有问题吗?我想使用功能进行读取和打印. 例如，当我输入n = 2时，m = 2且k = 2.对于矩阵的每个元素，我输入

I've been trying to implement a program in C, where i have two 2D arrays (created with pointers) and i have to multiply them (like multiplying matrices) and store the results to a third array. The dimensions of the arrays are given by the user. I've done the code but i'm getting wrong results. Is there something wrong with my formula? I wanted to do the reading and printing with functions. For example, when i input n = 2, m = 2 and k =2. For each element of the matrix i input

A(0)(0) = 1, A(0)(1) = 2, A(1)(0) = 3, A(1)(1) = 4 
and
B(0)(0) = 1,> B(0)(1) = 2, B(1)(0) = 3, B(1)(1) = 4.

输出应该是

C(0)(0) = 7, C(0)(1) = 10, C(1)(0) = 15 and C(1)(1) = 22.

相反，输出是

C(0)(0) = 7, C(0)(1) = 10, C(1)(0) = 33, C(1)(1) = 46.

我希望这不难理解，还不允许发布图像:(

I hope it's not hard to understand, i'm not allowed to post an image yet :(

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

int **read(int **x, int i, int j);
int **prod(int **x, int **y, int n, int m, int k);
void print(int **x, int r, int c);

int main()
{
int n, m, k, i, j;

printf("Give n, m and k: ");
scanf("%d%d%d", &n, &m, &k);

int **A, **B, **C;

A = (int **)malloc(n*sizeof(int *));
for (i = 0; i < n; i++)
    *(A+i) = (int *)malloc(k*sizeof(int));

B = (int **)malloc(k*sizeof(int *));
for (i = 0; i < k; i++)
    *(B+i) = (int *)malloc(m*sizeof(int));

C = (int **)malloc(n*sizeof(int *));
for (i = 0; i < n; i++)
    *(C+i) = (int *)malloc(m*sizeof(int)

for (i = 0; i < n; i++)
    for (j = 0; j < k; j++)
        A = read(A, i, j);

for (i = 0; i < k; i++)
    for (j = 0; j < m; j++)
        B = read(B, i, j);

C = prod(A, B, n, m, k);

print(C, n, m);
}

int **read(int **x, int i, int j)
{
printf("Give value to store in cell [%d][%d]: ", i, j);
scanf("%d", &x[i][j]);

return x;
}

int **prod(int **x, int **y, int n, int m, int k)
{
int i, j, l, sum;
int **res;

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

        for (l = 0; l < m; l++)
            sum = sum + (x[i][l]*y[l][j]);

        res[i][j] = sum;
    }

return res;
}

void print(int **x, int r, int c)
{
int i, j;

for (i = 0; i < r; i++)
{
    for (j = 0; j < c; j ++)
        printf("C[%d][%d]: %d\t", i, j, x[i][j]);

    printf("\n");
}
}

推荐答案

首先了解没有指针的矩阵

first understand the matrix without pointers

没有指针的矩阵乘法

int r1，r2，c1，c2;

int r1,r2,c1,c2;

int i，j，sum，k;

int i,j,sum,k;

void matrixmul()

void matrixmul()

{

printf("Getting values for matrices");

printf("\n---------------------------");

printf("\n Enter the row for Matrix A: ");

scanf("%d",&r1);

printf("\n Enter the column for Matrix A: ");

scanf("%d",&c1);

printf("\n Enter the row for Matrix B: ");

scanf("%d",&r2);

printf("\n Enter the column for Matrix A: ");

scanf("%d",&c2);

int a[r1][c1];

int b[r2][c2];

int c[r1][c2];

if (!((c1==r2)&&(r1==c2)))

{

    printf("Row column miss matching so cannot do matrix multiplication");

}

else

{

    printf("\n A Matrix");

    printf("\n ---------");

    for (i=0;i<r1;i++)

      {

          for(j=0;j<c1;j++)

          {

              printf("\n Enter %d and %d value: ",i,j);

              scanf("%d",&a[i][j]);

          }

      }

    printf("\n B Matrix");

    printf("\n ---------");

    for (i=0;i<r2;i++)

      {

          for(j=0;j<c2;j++)

          {

              printf("\n Enter %d and %d value: ",i,j);

              scanf("%d",&b[i][j]);

          }

      }
}

for(i=0;i<c1;i++)

   {

       for(j=0;j<r2;j++)

       {

           for (k=0;k<r2;k++)

           {

               sum = sum + a[i][k]*b[k][j];

            }

            c[i][j] = sum;

            sum = 0;
       }

   }

printf("\nResultant Matrix");

printf("\n----------------\n");

for(i=0;i<r1;i++)

{

    for(j=0;j<c2;j++)

    {

        printf("%d\t",c[i][j]);

    }

    printf("\n");

}

}

void main()

void main()

{

matrixmul();

}

输出

输入矩阵A的行:2

输入矩阵A的列:3

输入矩阵B的行:3

输入矩阵B的列:2

矩阵

输入0和0值:1

输入0和1值:2

输入0和2值:3

输入1和0值:3

输入1和1值:2

输入1和2值:11

B矩阵

输入0和0值:4

输入0和1值:5

输入1和0值:7

输入1和1值:8

输入1和0值:9

输入1和1的值:10

Enter the 1 and 1 value:10

结果矩阵

45 51
125141

45 51
125 141

带有指针的矩阵-在此程序中，使用指针变量声明2D并获取2D数组的值并显示值

Matrix with pointers-in this program 2D is declared using pointer variable and get the values for the 2D array and display the values

void main()

void main()

{

int r = 2, c = 3, i, j, count;

int *a[r];//while declaring the array as pointer its number of rows are specified

for (i=0; i<r; i++)

{

    a[i] = (int *)malloc(c * sizeof(int));//Allocate memory space for each row for 'c' columns

}


count = 0;

for (i = 0; i < r; i++)

{

    for (j = 0; j < c; j++)

        {

            scanf("%d",&a[i][j] ); 

        }   

}


for (i = 0; i < r; i++)

{

    for (j = 0; j < c; j++)

        {


           printf("%d\t ", a[i][j]);

        }

    printf("\n");

}

}

输出

3

2

1

1

2

3

3 2 1

1 2 3

以下是使用指针的最终完整矩阵乘法

The following is the final complete matrix multiplication using pointers

void input()

void input()

{

int r1, c1,r2,c2, i, j,k,sum;

printf("\n Enter the row for Matrix A: ");

scanf("%d",&r1);

printf("\n Enter the column for Matrix A: ");

scanf("%d",&c1);

printf("\n Enter the row for Matrix B: ");

scanf("%d",&r2);

printf("\n Enter the column for Matrix A: ");

scanf("%d",&c2);

if (!((c1==r2)&&(r1==c2)))

{

     printf("Row column miss matching so cannot do matrix multiplication");

}

else

{
    int *a[r1];

    int *b[r2];
    int *c[r1];

    for (i=0; i<r1; i++)

    {

        a[i] = (int *)malloc(c1 * sizeof(int));

    }

    for (i=0; i<r2; i++)

    {

        b[i] = (int *)malloc(c2 * sizeof(int));


    }

    for (i=0; i<r1; i++)

    {

        c[i] = (int *)malloc(c2 * sizeof(int));

    }

    printf("\n A Matrix");
    printf("\n ----------");

    for (i = 0; i < r1; i++)

        {

            for (j = 0; j < c1; j++)

                 {

                    printf("\n Enter the %d and %d value:",i,j);

                    scanf("%d",&a[i][j] ); 

                }   

        }
        printf("\n B Matrix");
        printf("\n ----------");

        for (i = 0; i < r2; i++)

            {

                for (j = 0; j < c2; j++)

                    {

                        printf("\n Enter the %d and %d value:",i,j);

                         scanf("%d",&b[i][j]); 

                      } 

                }

        for(i=0;i<c1;i++)

           {

                for(j=0;j<r2;j++)

                    {

                        for (k=0;k<r2;k++)

                            {

                                sum = sum + a[i][k]*b[k][j];

                            }

                         c[i][j] = sum;

                        sum = 0;
                     }

            }

        printf("\n Resultant Matrix");

        printf("\n------------------");

        for (i = 0; i < r1; i++)

            {

                for (j = 0; j < c2; j++)

                    {

                        printf("%d\t ", c[i][j]);

                     }

                printf("\n");

             }

    }

}

void main() {

void main() {

input();

}

输出

输入矩阵A的行:2

输入矩阵A的列:3

输入矩阵B的行:3

输入矩阵B的列:2

矩阵

输入0和0值:1

输入0和1值:2

输入0和2值:3

输入1和0值:3

输入1和1值:2

输入1和2值:11

B矩阵

输入0和0值:4

输入0和1值:5

输入1和0值:7

输入1和1值:8

输入2和0值:9

输入2和1的值:10

Enter the 2 and 1 value:10

结果矩阵

45 51
125141

45 51
125 141

一个个地运行上面的程序，了解它，然后将最终程序转换为所需的功能，谢谢

run the above program one by one understand it then convert the final program into function as per your required,Thank you

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