Question

要求用户在控制台输入整数“n”。您的程序将在0到1023（含）之间生成两个n x n个随机整数矩阵，然后将它们相乘，相加和相减。矩阵将表示为二维阵列。您的程序应显示每个矩阵运算的结果（乘法，减法和加法）。

（对于减法部分，我是否需要第三个空矩阵来表示？）没有得到它。

以下是我的代码。运行不顺利。我的数组乘法似乎是正确的，然而，在我点击编译并运行后，它只是在用户输入输入后陷入困境。这是怎么用C编写矩阵的？谢谢，

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

int main() {
    int number;

    //ask for interger
    printf("\nEnter and interger: ");
    scanf("%d", &number);

    int Array1[number][number];
    int Array2[number][number];
    int Array3[number][number];
    int i,j, r;

    //populate array
    for(i = 0; i < number; i++ ) {
            for(i = 0; j < number; j++ ) {
                   r = rand() % 1023;
                   Array1[i][j] = r;
                   Array2[i][j] = r;
            }
    }

    //calutate array
    //multiplication
    for(i = 0; i < number; i++ ) {
            for(i = 0; j < number; j++ ) {
                   Array3[i][j] = 
                       Array1[i][j] * Array2[i][j] + Array1[i][j+1] * Array2[i+1][j];
            }
    }

    return 0;

}

Answer 1

使用VLA简化内存管理时，至少应检查用户输入的大小是否有效。

您的代码中也存在一些错误，但从数学角度来看，最糟糕的是您如何计算两个矩阵的乘积。考虑 C = AB ，正确的公式为：

C _i，j = Σ_k A _i，k B _{k ，J}的

尝试这段代码，我评论并修复了我发现的问题：

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

// Declarations of functions which calculate the matrix product and print out
// the results. I assume square matrices stored as variable length arrays
void matrix_product( int width, int a[][width], int b[][width], int c[][width] );
void print_array( char *msg, int width, int arr[][width] );

int main() {
    int number = 0, rs;

    //ask for an integer
    printf("\nEnter and interger: ");
    // It's good practice to check the return value of scanf
    while ( (rs = scanf("%d", &number)) != 1  ||  number < 1 ) {
        if ( rs == EOF ) {
            printf("\nFatal Error: No input\n");
            return EXIT_FAILURE;
        }
        printf("\nPlease, enter a positive integer: ");
        // clear input buffer reading whatever left on the line till a newline
        // "[^\n]", without storing it anywhere "*"
        scanf("%*[^\n]");  
    }

    int Array1[number][number];
    int Array2[number][number];
    int Array3[number][number];
    int i, j, r;

    // populate array
    srand(time(NULL));
    for( i = 0; i < number; i++ ) {
        for( j = 0; j < number; j++ ) {
        //  ^^^ it was i in your code    
            r = rand() % 1024;
            // note that 1024 % 1024 = 0  while  1023 % 1024 = 1023
            Array1[i][j] = r;
            // I guess that the two matrices are intended to be different
            r = rand() % 1024;
            Array2[i][j] = r;
        }
    }

    // print the original arrays
    print_array("\nArray 1:", number, Array1);
    print_array("\nArray 2:", number, Array2);

    // print the result of calculations
    matrix_product(number, Array1, Array2, Array3);
    print_array("\nArray 3, result of multiplication:", number, Array3);

    return EXIT_SUCCESS;

}

// print out the elements of a square matrix stored as a 2D VLA
// preceded by an informative string
void print_array( char *msg, int width, int arr[][width] ) {
    int i, j;
    printf("%s\n", msg);
    for( i = 0; i < width; i++ ) {
        for( j = 0; j < width; j++ ) {
            printf("%8d", arr[i][j]);
        }
        printf("\n");
    }   
}

// Calculate the matrix product c = a * b assuming square matrices
// stored as 2D VLA
void matrix_product( int width, int a[][width], int b[][width], int c[][width] ) {
    int i, j, k, sum;
    for( i = 0; i < width; ++i ) {
        for( j = 0; j < width; ++j ) {
            sum = 0;
            for ( k = 0; k < width; ++k ) {
                sum += a[i][k] * b[k][j];
            }
            c[i][j] = sum;
        }
    }
}

C矩阵运算

1 个答案: