动态分配2d阵列 [英] Dynamically allocated 2d array

问题描述

我正在尝试创建一个2d数组，特别是有向图的邻接矩阵。
我从来没有尝试使用动态内存分配，之前我打了一个陷阱。
这是代码：

  int n，i; 
 printf（节点数为：）; 
 scanf（％d，& n）; 
 
 int ** array = malloc（n * sizeof（int *））; 
 for（i = 0; i  array [i] = malloc（n * sizeof（int））; 
 
 printf（边数为：）; 
 scanf（％d，& m）; 
 
 int x，y; （i = 0; i< m; i ++）
 {
 scanf（％d％d，& x，& y） 
 array [x] [y] = 1; 
} 
  

一旦完成了所有边缘的输入，程序将停止工作并抛出

编辑：houssam发现我的错误。第一个for应该从1到n。当我输入1 6作为边缘，程序崩溃，因为只有节点0-5。感谢您的发现。这样一个简单的错误.....

解决方案

您可能输入错误的价值 x 或 y ，其值应小于 n ，大于或等于 0 ：

  #include< stdlib.h> 
 #include< stdio.h> 
 
 int main（）
 {
 int m，n，i; 
 printf（节点数为：）; 
 scanf（％d，& n）; 
 
 int ** array = malloc（n * sizeof（int *））; 
 for（i = 0; i  array [i] = malloc（n * sizeof（int））; 
 
 printf（边数为：）; 
 scanf（％d，& m）; 
 
 int x，y; （i = 0; i< m; i ++）
 {
 scanf（％d％d，& x，& y） 
 if（x> = 0&&&> = 0&& x< n&& y< n）//这里
 array [x] ] = 1; 
 else 
 printf（你的值中的错误x =％d y =％d\\\
，x，y）; 
} 
 return 0; 
} 
  

编辑＃1：

基于用户的评论： M Oehm ，以检查 scanf ，请参阅：

scanf失败为什么？，我可以重构代码如：

  #include< stdlib.h> 
 #include< stdio.h> 
 
 int main（）
 {
 int m，n，i; 
 printf（节点数为：）; 
 scanf（％d，& n）; 
 
 int ** array = malloc（n * sizeof（int *））; 
 for（i = 0; i  array [i] = malloc（n * sizeof（int））; 
 
 printf（边数为：）; 
 scanf（％d，& m）; 
 
 int x，y; （i = 0; i< m; i ++）
 
 {
 // int num_read = scanf（％d％d，& x，& y） 
 if（scanf（％d％d，& x，& y）< 2）
 {
 printf（请输入有效的两个整数... \\ ）; 
 while（getchar（）！='\\\
'）; //读取输入流中的所有字符。 
} 
 else if（x> = 0&&&> = 0&& x< n&&< n）// here 
 {
 array [x] [y] = 1; 
 printf（array [％d] [％d] = 1; \\\
，x，y）; 
 
} 
 else 
 printf（你的值中的错误x =％d y =％d\\\
，x，y）; 
} 
 return 0; 
} 
  

I'm trying to create a 2d array, specifically an adjacency matrix for directed graphs. I've never tried this with dynamic memory allocation before and I've hit a snag. Here's the code:

int n, i;
printf("Number of nodes is: ");
scanf("%d", &n);

int ** array = malloc(n * sizeof(int*));
for(i = 0; i < n; i++)
    array[i] = malloc(n * sizeof(int));

printf("Number of edges is: ");
scanf("%d", &m);

int x, y;
for(i=0;i<m;i++)
{
    scanf("%d %d", &x, &y);
    array[x][y]=1;
}

As soon as I finish entering all the edges, the program stops working and throws the usual "exe has stopped working".

Where's the problem?

EDIT: houssam spotted my error. The first "for" should have gone from 1 to n. When I entered 1 6 as an edge, the program crashed because there were only nodes 0-5. Thanks for spotting that. Such a simple mistake.....

解决方案

you may entered wrong values for x or y , their values should be less than n and greater than or equal to 0:

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

int main()
{
    int m ,n, i;
    printf("Number of nodes is: ");
    scanf("%d", &n);

    int ** array = malloc(n * sizeof(int*));
    for(i = 0; i < n; i++)
        array[i] = malloc(n * sizeof(int));

    printf("Number of edges is: ");
    scanf("%d", &m);

    int x, y;
    for(i=0;i<m;i++)
    {
        scanf("%d %d", &x, &y);
        if (x >=0 && y >= 0 && x < n && y < n) // here
            array[x][y]=1;
        else
            printf("error in your values x=%d y=%d\n" , x , y);
    }
    return 0;
}

EDIT #1:
based on comment from user : M Oehm to check the return value of scanf , see:
scanf fails why? , I can refactor the code to be like:

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

int main()
{
    int m ,n, i;
    printf("Number of nodes is: ");
    scanf("%d", &n);

    int ** array = malloc(n * sizeof(int*));
    for(i = 0; i < n; i++)
        array[i] = malloc(n * sizeof(int));

    printf("Number of edges is: ");
    scanf("%d", &m);

    int x, y;
    for(i=0;i<m;i++)
    {
        //int num_read = scanf("%d %d", &x, &y);
        if(scanf("%d %d", &x, &y) < 2)
        {
            printf("please enter valid two integers..\n");          
            while (getchar() != '\n'); // read all characters in the input stream.
        }
        else if (x >=0 && y >= 0 && x < n && y < n) // here
        {
            array[x][y]=1;
            printf("array[%d][%d]=1;\n" , x, y);

        }
        else
            printf("error in your values x=%d y=%d\n" , x , y);
    }
    return 0;
}

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