使用函数检查相等的矩阵

Question

我正在尝试通过函数检查两个矩阵是否相等。但我很困惑如何为函数EqualMatrices 编写参数。由于逻辑是比较相等或行和列如何在函数参数中传递这样的数组。

#include <stdio.h> 
    
    void EqualMatrices(int A[][10], int B[][10]) // how to write arguments for this function
    {
        /*  Comparing two matrices for equality: this is the logic */ 

        if (row1 == row2 && column1 == column2)

        {

            printf("Matrices can be compared \n");

            for (i = 0; i < row1; i++) 

            {

                for (j = 0; j < column2; j++)

                {

                    if (a[i][j] != b[i][j])

                    {

                        flag = 0;

                        break;

                    }

                 }

            }

        }

        else 

        {

            printf(" Cannot be compared\n");

            exit(1);

        }

 

        if (flag == 1)

            printf("Two matrices are equal \n");

        else

            printf("But, two matrices are not equal \n");
    }

void main() 
    { 

        int a[10][10], b[10][10];

        int i, j, row1, column1, row2, column2, flag = 1;

 

        printf("Enter the order of the matrix A \n");

        scanf("%d %d", &row1, &column1);
 

        printf("Enter the order of the matrix B \n");

        scanf("%d %d", &row2, &column2);
        
        
            printf("Enter the elements of matrix a \n");

            for (i = 0; i < row1; i++)
    
            {
    
                for (j = 0; j < column1; j++) 
    
                {
    
                    scanf("%d", &a[i][j]);
    
                }
    
            }

        printf("Enter the elements of matrix b \n");

        for (i = 0; i < row2; i++) 

        {

            for (j = 0; j < column2; j++)

            {

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

            }

        }

 

        printf("MATRIX a is \n");

        for (i = 0; i < row1; i++)

        {

            for (j = 0; j < column1; j++) 

            {

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

            }

            printf("\n");

        }

 

        printf("MATRIX b is \n");

        for (i = 0; i < row2; i++)

        {

            for (j = 0; j < column2; j++) 

            {

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

            }

            printf("\n");

 

        }
        
        EqualMatrices(a, b);

        return 0; 

    }

Answer 1

希望这会奏效。我没有使用指针就做到了。

 #include <stdio.h> 
    
    void EqualMatrices(int A[][10], int B[][10], int row1, int column1, int row2, int column2)
    {
        /*  Comparing two matrices for equality */ 
        int i,j, flag = 1;

        if (row1 == row2 && column1 == column2)
        {
            printf("Matrices can be compared \n");
            for (i = 0; i < row1; i++) 
            {
                for (j = 0; j < column2; j++)
                {
                    if (A[i][j] != B[i][j])
                    {
                        flag = 0;
                        break;
                    }
                 }
            }
        }
        else 
        {
            printf("\n Cannot be compared\n");
            exit(1);
        }
        if (flag == 1)
        {
            printf("\n Two matrices are equal \n");
        }
        else
        {
            printf("\n Two matrices are not equal \n");
        }   
    }

    int main() 
    { 
        int a[10][10], b[10][10];
        int i, j, row1, column1, row2, column2;
        
        printf ("Enter the order of the matrix A (mxn):\n");
        printf ("Row of matrix (mxn): ");
        scanf ("%d", &row1);
        printf ("Column of matrix (mxn): ");
        scanf ("%d", &column1);
        
        printf ("Enter the order of the matrix B (mxn):\n");
        printf ("Row of matrix (mxn): ");
        scanf ("%d", &row2);
        printf ("Column of matrix (mxn): ");
        scanf ("%d", &column2);

        
            printf("Enter the elements of matrix a \n");

            for (i = 0; i < row1; i++)  
            {   
                for (j = 0; j < column1; j++)   
                {
                    printf("Enter a[%d][%d]: ",i,j);    
                    scanf("%d", &a[i][j]);  
                }   
            }

        printf("Enter the elements of matrix b \n");

        for (i = 0; i < row2; i++)
        {
            for (j = 0; j < column2; j++)
            {
                printf("Enter b[%d][%d]: ",i,j);    
                scanf("%d", &b[i][j]);
            }
        }
        printf("\n MATRIX a is \n");
        
        for (i = 0; i < row1; i++)
        {
            printf("\n");
            for (j = 0; j < column1; j++) 
            {
                printf("%d\t", a[i][j]);
            }
        }

        printf("\n MATRIX b is \n");
        
        for (i = 0; i < row2; i++)
        {
            printf("\n");
            for (j = 0; j < column2; j++) 
            {
                printf("%d\t", b[i][j]);
            }
            printf("\n");
        }
        
        EqualMatrices(a, b, row1, column1, row2, column2);
        
        return 0; 
    }

Answer 2

这个问题与那里已经回答的更一般的问题有关： How are multi-dimensional arrays formatted in memory? / How to pass a 2D array by pointer in C?

Here，尝试创建简单的矩阵比较函数，说明如何直接使用普通数组：

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

/* VARIANT 0: compare elements of arrays in a nested loop
   -> no obvious advantages over VARIANT 1 */
bool matrix_equal0(
    const int *a, const int *b, unsigned int rows, unsigned int columns)
{
    for (unsigned int i = 0; i < rows; i++) {
        for (unsigned int j = 0; j < columns; j++) {
            if (a[i * columns + j] != b[i * columns + j]) {
                return false;
            }
         }
    }
    return true;
}

/* VARIANT 1: compare elements of matrices one by one,
   simpler and more performant than VARIANT 0 */
bool matrix_equal1(
    const int *a, const int *b, unsigned int rows, unsigned int columns)
{
    unsigned int i = rows * columns;
    while (i--) {
        if (*a++ != *b++) {
            return false;
        }
    }
    return true;
}

/* VARIANT 2:
   compare both arrays as a chunk of memory (possible but not recommended) */
bool matrix_equal2(
    const int *a, const int *b, unsigned int rows, unsigned int columns)
{
    return (memcmp(a, b, rows * columns * sizeof(int)) == 0);
}

int main(void)
{
    const int mat0[3][3] = {
        { 1, 2, 3, },
        { 4, 5, 6, },
        { 7, 8, 9, },
    };
    const int mat1[3][3] = {
        { 1, 2, 3, },
        { 4, 5, 6, },
        { 7, 8, 9, },
    };
    const int mat2[3][3] = {
        { 1, 2, 3, },
        { 4, 5, 6, },
        { 7, 8, 9999, },
    };
    if (matrix_equal0(&mat0[0][0], &mat1[0][0], 3, 3)) {
        printf("VARIANT 0: mat0 and mat1 are equal!\n");
    }
    if (matrix_equal0(&mat0[0][0], &mat2[0][0], 3, 3) == false) {
        printf("VARIANT 0: mat0 and mat2 are NOT equal!\n");
    }

    if (matrix_equal1(&mat0[0][0], &mat1[0][0], 3, 3)) {
        printf("VARIANT 1: mat0 and mat1 are equal!\n");
    }
    if (matrix_equal1(&mat0[0][0], &mat2[0][0], 3, 3) == false) {
        printf("VARIANT 1: mat0 and mat2 are NOT equal!\n");
    }

    if (matrix_equal2(&mat0[0][0], &mat1[0][0], 3, 3)) {
        printf("VARIANT 2: mat0 and mat1 are equal!\n");
    }
    if (matrix_equal2(&mat0[0][0], &mat2[0][0], 3, 3) == false) {
        printf("VARIANT 2: mat0 and mat2 are NOT equal!\n");
    }
    return 0;
}

输出：

VARIANT 0: mat0 and mat1 are equal!
VARIANT 0: mat0 and mat2 are NOT equal!
VARIANT 1: mat0 and mat1 are equal!
VARIANT 1: mat0 and mat2 are NOT equal!
VARIANT 2: mat0 and mat1 are equal!
VARIANT 2: mat0 and mat2 are NOT equal!

注意：当二维矩阵被视为普通数组时，应注意。列和行如何与数组的一维内存布局相关可能会造成混淆并导致错误。 - Here a quick example，说明了这种关系：

#include <stdio.h>

/* print the elements of a 2D matrix */
void matrix_print(const int *mat, unsigned int rows, unsigned int columns)
{
    for (unsigned int i = 0; i < rows; i++) {
        for (unsigned int j = 0; j < columns; j++) {
            printf("%d ", mat[i * columns + j]);
        }
        printf("\n");
    }
    printf("\n");
}

int main(void)
{
    const int mat[2][3] = {
        { 1, 2, 3, },
        { 4, 5, 6, },
    };
    matrix_print(&mat[0][0], 2, 3);
    return 0;
}

输出：

1 2 3 
4 5 6 

---

如果您的项目需要大量的矩阵数学，那么使用自定义类型以使其更易于处理并构建一组库式的实用函数可能是有意义的。 - Here 一个示例，说明在这种情况下比较函数的外观：

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

/* VARIANT 3: instead of using a 2D array to represent a matrix, 
              a custom type is used. That makes it easier/nicer to pass
              matrices around various places in your code. */
typedef struct {
    unsigned int rows;
    unsigned int columns;
    int *elements;
} matrix;

matrix matrix_new(unsigned int rows, unsigned int columns)
{
    matrix mat;
    mat.rows = rows;
    mat.columns = columns;
    mat.elements = calloc(rows * columns, sizeof(int)); /* zero matrix */
    return mat;
}

bool matrix_clear(matrix *mat)
{
    if (mat == NULL)
        return false;
    mat->rows = 0;
    mat->columns = 0;
    free(mat->elements);
    mat->elements = NULL;
    return true;
}

bool matrix_elem(matrix *mat, unsigned int row, unsigned int column, int val)
{
    if (mat == NULL)
        return false;
    if (row >= mat->rows || column >= mat->columns)
        return false;
    if (mat->elements == NULL)
        return false;
    mat->elements[row * mat->columns + column] = val;
    return true;
}

bool matrix_compare(const matrix *mat0, const matrix *mat1)
{
    if (mat0 == NULL || mat1 == NULL)
        return false;
    if (mat0->rows != mat1->rows)
        return false;
    if (mat0->columns != mat1->columns)
        return false;
    int *a = mat0->elements;
    int *b = mat1->elements;
    unsigned int i = mat0->rows * mat0->columns;
    while (i--) {
        if (*a++ != *b++) {
            return false;
        }
    }
    return true;
}

int main(void)
{
    matrix mat3 = matrix_new(3, 3);
    matrix mat4 = matrix_new(3, 3);
    matrix mat5 = matrix_new(3, 3);
    matrix_elem(&mat3, 1, 1, 12);
    matrix_elem(&mat4, 1, 1, 12);
    matrix_elem(&mat5, 1, 1, 13);
    //matrix_copy_from_2d_array(&mat6, mat0); // todo
    if (matrix_compare(&mat3, &mat4)) {
        printf("VARIANT 3: mat3 and mat4 are equal!\n");
    }
    if (matrix_compare(&mat3, &mat5) == false) {
        printf("VARIANT 3: mat3 and mat5 are NOT equal!\n");
    }
    matrix_clear(&mat3);
    matrix_clear(&mat4);
    matrix_clear(&mat5);
    
    return 0;
}

输出：

VARIANT 3: mat3 and mat4 are equal!
VARIANT 3: mat3 and mat5 are NOT equal!