c - 指针使用不正确导致输出错误？

Question

我有一个已完全填充的结构。这个名字是strdup'd。我想要做的是，创建一个指针数组，其中数组中的元素指向该 strdup 的变量。

这是我尝试过的:

struct Student{
        char *name;
        int age;
}Student;

void storeAdrress(const Student *student1,char **arrayOfPointers){
    arrayOfPointers[0] = student1->name;    

    printf("Struct variable: %p",student1->name);
    printf("Pointer variable: %p",arrayOfPointers[0]);
}

int main(void){
    Student *student1 = calloc(1,sizeof(Student));
    student1->name = strdup("Tom");

    char *arrayOfPointers[100] = {0};

    storeAdrress(student1,arrayOfPointers);

    free(student1->name);
    free(student1);

    return(0);
}

这是我的输出:

Struct variable: (0x5502a80)
Pointer variable: (nil)

我的目标是让arrayOfPointers[0]获得名称的地址。这显然不起作用，因为 arrayOfPointers[0] 没有存储指针位置。

有人可以告诉我我做错了什么吗？

Answer 1

Weather Vane 击败了我，但这里是对您的代码的额外更正，没有使用 typedef，并添加了一些错误检查:

#define _POSIX_C_SOURCE 200809L

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

struct Student{
    char *name;
    int age;
};

void storeAddress(struct Student * stdt, char ** pArray, size_t index)
{
    pArray[index] = stdt->name;    
    printf("Struct variable: %p\n", (void *) stdt->name);
    printf("Pointer variable: %p\n", (void *) pArray[index]);
}

int main(void)
{
    struct Student *student1 = calloc(1,sizeof(*student1));
    if ( !student1 ) {
        perror("couldn't allocate memory for struct Student");
        exit(EXIT_FAILURE);
    }

    student1->name = strdup("Tom");
    if ( !student1->name ) {
        perror("couldn't allocate memory for student name");
        exit(EXIT_FAILURE);
    }

    char *arrayOfPointers[100] = {0};

    storeAddress(student1, arrayOfPointers, 0);

    free(student1->name);
    free(student1);

    return 0;
}

输出:

Paul@Pauls-iMac:~/Documents/src/sandbox/names$ ./names
Struct variable: 0x7fd632403210
Pointer variable: 0x7fd632403210
Paul@Pauls-iMac:~/Documents/src/sandbox/names$