Question

这里是简单的代码，只读两个矩阵一个是3 * 3维，另一个是3 * 1维。在打印第一个矩阵A [3] [3]时，矩阵的最后一个元素在void printarray(double **A, int n )函数中打印为零。

在我的代码下面：

#include <stdio.h>
#include<malloc.h>

void printarray(double **A, int n );

void main(){
double **A;
int n = 3;
int row,col;
double *b;

A = (double **) malloc(n * sizeof(double**));

for (row = 1; row<= n; row++) {
    A[row] = (double *) malloc(n * sizeof(double));
    }

// Initialize each element.
for (row = 1; row<= n; row++) {
    for (col = 1; col<= n; col++) {
        printf("A[%d][%d]= %u \t",row,col,&A[row][col]);
        scanf("%lf",&A[row][col]);  // or whatever value you want
    }
}

//print A
printf("\n...........array in main.................\n");
for (row = 1; row<= n; row++) {
    for (col = 1; col<= n; col++) {
        printf("A[%d][%d]=%u \t %lf",row,col,&A[row][col],A[row][col]);
        printf("\n");
    }
}

b = (double *) malloc(n * sizeof(double));

printf("\n enter the elemet of b \n"); // Initialize each element.
for (row = 1; row<= n; row++){
    printf("address=%u \t",&b[row]);
    printf("b[%d]=",row);
    scanf("%lf",&b[row]);
    printf("\n");
}

printarray((double **)A, n );

}// Print it

void printarray(double **A, int n ){
    int i;
    int j;
    printf("\n.....print a.............\n");
    for( j = 1; j <= n; j++ ){
        for( i = 1; i <= n; i ++){
            printf("A[%d][%d]= %u \t",j,i,&A[j][i]);
            printf( "%lf ", A[j][i] );
        }
        printf( "\n" );
    }
}

Answer 1

问题之一来自数组的索引。数组索引从0开始。

这意味着，为了遍历您的数组，您需要for循环从0开始，直到n-1：

for (int row=0; row<n;++row) {/*...*/}

当使用双指针传递2d矩阵到打印功能时，它打印矩阵零的最后一个元素

1 个答案: