如何使结构体在C用户定义的数组 [英] How to make a user defined array of struct in C

问题描述

我想用户定义数组的大小在程序启动时，我目前有：

I would like the user to define the size of the array when the program starts, I currently have:

#define SIZE 10
typedef struct node{
    int data;
    struct node *next;
} node;

    struct ko {
    struct node *first;
    struct node *last;
} ;

struct ko array[SIZE];

这不过的工作，我想删除的#define SIZE ，让大小是用户定义的值，所以在主函数中我有：

This works, however, I would like to remove the #define SIZE, and let SIZE be a value that the user defines, so in the main function i have:

int SIZE;
printf("enter array size");
scanf("%d", &SIZE);

我怎么可以得到价值的阵列？

how can I get that value to the array?

编辑：
现在我已经中.h文件中以下内容：

now i have the following in the .h file:

    typedef struct node{
    int data;
    struct node *next;
    } node;

    struct ko {
    struct node *first;
    struct node *last;
    } ;

struct ko *array;
int size;

和这main.c文件：

and this in the main.c file:

printf("size of array: ");
scanf("%d", &size);
array = malloc(sizeof(struct ko) * size);

如若这项工作？它没有程序崩溃，但我不知道，如果问题
在这里，或者其他地方的计划...

Should this work? It doesn't the program crashes but I don't know if the problem is here or, elsewhere in the program...

推荐答案

而不是结构こ阵列[SIZE]; ，动态分配的：

struct ko *array;
array = malloc(sizeof(struct ko) * SIZE);

确认释放它，一旦你用它做：

Make sure to free it once you're done with it :

free(array);

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