应该 return 结构指针的函数

Question

我正在研究结构指针及其在函数中的用法。在这里，我有一个问题。下面的代码没有打印我希望它打印的 ID、名称和分数。我花了几个小时试图弄清楚这一点。谁能通俗易懂的解释一下？

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct examinee
{
    char ID[8];
    char name[32];
    int score;
};
 
struct examinee* init(char ID[8], char name[32], int score){
    struct examinee* instance; 
    instance = (struct examinee*)malloc(sizeof(struct examinee));
    strcpy(instance->ID, ID);
    strcpy(instance->name, name);
    instance->score;
    return  instance;
};

int main(int argc, char *argv[])
{
    struct examinee examinee_1;
    struct examinee* examinee_1_ptr = init("IO760","Abe",75);
    examinee_1_ptr= &examinee_1;
    printf("examinee's information:\nID: %s\nname: %s\nscore: %d\n", examinee_1_ptr->ID, examinee_1_ptr->name, examinee_1_ptr->score);
    return 0;
}

Answer 1

在这一行

struct examinee* examinee_1_ptr = init("IO760","Abe",75);

您执行该函数并将结果存储在 examinee_1_ptr 变量中。

但是这两行

struct examinee examinee_1; 
examinee_1_ptr= &examinee_1;

创建一个空的 examinee 实例并用该空实例的地址覆盖指针。此时，您在 init() 中创建的那个丢失了，因为没有任何东西再指向它的地址，所以您无法访问它。去掉这两行就可以了。

printf("examinee's information:\nID: %s\nname: %s\nscore: %d\n", examinee_1_ptr->ID, examinee_1_ptr->name, examinee_1_ptr->score);

注意这里你只是通过指针访问对象。首先不需要 examinee_1 变量。

应该 return 结构指针的函数

Function that should return structure pointer

c

function

pointers

structure