将整数复制到指针结构变量

Question

我正在尝试为指针结构变量赋值。 这就是我想要做的 -

struct node
    {
        char name[31];
        int probability[3];
        struct node *ptr;
    };
    typedef struct node NODE;

NODE *head;

   head  = (NODE *)malloc(sizeof(NODE));
   head->probability[0] = atoi(strtok (buf," -"));//Doesn't assingn
   head->probability[1] = atoi(strtok(NULL," -"));//Doesn't assingn
   head->probability[2] = atoi(strtok(NULL," -"));//Doesn't assingn

这里的“buf”是一个包含“5 10 0”格式值的字符串。但是上面的代码没有给 head->probability 赋值。 请给我一个解决方案。

Answer 1

它对我来说很好用。这是我的代码

// FILE: test_struct.c

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

#define NAME_LEN 31
#define PROB_LEN 3

struct node {
    char name[NAME_LEN];
    int probability[PROB_LEN];
    struct node *ptr;
};

typedef struct node NODE;

int main() 
{
   char buf[] = "5 10 0";
   NODE *head;
   int i;

   head = (NODE *)malloc(sizeof(NODE));
   assert(head);

   // add probabilities
   for (i=0; i < PROB_LEN; i++)
       head->probability[i] = atoi(strtok ((i == 0) ? buf : NULL," -"));

   // print probabilities
   for (i = 0; i < PROB_LEN; i++)
       printf("head->probability[%d] = %d\n", i, head->probability[i]);

   free(head);
   return 0;
}

编译与执行

gcc test_struct.c -o ts
./ts

打印结果

head->probability[0] = 5
head->probability[1] = 10
head->probability[2] = 0