在C中具有指针作为成员的结构的memcpy

Question

如果这是一个愚蠢的问题,请不要投票给我,我只是想了解.

我有一个结构,有一些指针作为成员,我正在尝试做memcpy,我有人建议我不应该在这种情况下使用memcpy作为memcpy做一个浅拷贝(意思是它复制指针)相当深的拷贝(意味着复制什么指针)指向).

但我不确定为什么它在以下程序中没有任何区别:请查看代码和输出,并解释为什么在这种情况下它不是浅层副本？

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

struct student {
    char *username;
    char *id;
    int roll;
};

void print_struct(struct student *);
void print_struct_addr(struct student *);
void changeme(struct student *);

int main (void) {
    struct student *student1;
    char *name = "ram";
    char *id = "200ABCD";
    int roll = 34;

    student1 = (struct student *)malloc(sizeof(struct student));
    student1->username = name; 
    student1->id = id;
    student1->roll = roll; 
    print_struct_addr(student1);
    print_struct(student1);
    changeme(student1);
    print_struct(student1);
    print_struct_addr(student1);
    return 0;
}

void print_struct(struct student *s) {
    printf("Name: %s\n", s->username);
    printf("Id: %s\n", s->id); 
    printf("R.No: %d\n", s->roll);
    return; 
}

void print_struct_addr(struct student *s) {
    printf("Addr(Name): %x\n", &s->username);
    printf("Addr(Id): %x\n", &s->id);
    printf("Addr(R.No): %x\n", &s->roll);
    return;
}

void changeme(struct student *s) {
    struct student *student2;
    student2->username = "someone";
    student2->id = "200EFGH";
    student2->roll = 35;
    print_struct_addr(student2);
    memcpy(s, student2, sizeof(struct student));
    student2->username = "somebodyelse";
    return;
}

输出:

Addr(Name): 9b72008
Addr(Id): 9b7200c
Addr(R.No): 9b72010
Name: ram
Id: 200ABCD
R.No: 34
Addr(Name): fa163c
Addr(Id): fa1640
Addr(R.No): fa1644
Name: someone
Id: 200EFGH
R.No: 35
Addr(Name): 9b72008
Addr(Id): 9b7200c
Addr(R.No): 9b72010

如果memcpy做了浅拷贝,那么student1-> username怎么不是"somebodyelse".

请解释在哪种情况下,这段代码可能会产生问题,我想在main()中的changeme()调用之后在student1中找到student2信息,之后应该可以使用这个修改过的student1数据.

我被建议不要在这里使用memcpy(),但似乎工作正常.

谢谢

这是修改后的代码:但我仍然没有看到浅拷贝的概念:

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

struct student {
    char *username;
    char *id;
    int roll;
};

void print_struct(struct student *);
void print_struct_addr(struct student *);
void changeme(struct student *);

int main (void) {
    struct student *student1;
    char *name = "ram";
    char *id = "200ABCD";
    int roll = 34;

    student1 = malloc(sizeof(*student1));
    student1->username = name; 
    student1->id = id;
    student1->roll = roll; 
    print_struct_addr(student1);
    print_struct(student1);
    changeme(student1);
    print_struct(student1);
    print_struct_addr(student1);
    return 0;
}

void print_struct(struct student *s) {
    printf("Name: %s\n", s->username);
    printf("Id: %s\n", s->id); 
    printf("R.No: %d\n", s->roll);
    return; 
}

void print_struct_addr(struct student *s) {
    printf("Addr(Name): %x\n", &s->username);
    printf("Addr(Id): %x\n", &s->id);
    printf("Addr(R.No): %x\n", &s->roll);
    return;
}

void changeme(struct student *s) {
    struct student *student2;
    student2 = malloc(sizeof(*s));
    student2->username = strdup("someone");
    student2->id = strdup("200EFGH");
    student2->roll = 35;
    print_struct_addr(student2);
    memcpy(s, student2, sizeof(struct student));
    student2->username = strdup("somebodyelse");
    free(student2);
    return;
}

Answer 1

这个:

struct student *student2;
student2->username = "someone";
student2->id = "200EFGH";
student2->roll = 35;

写入未分配的内存,调用未定义的行为.student2在写之前,你需要确保指向某个有效的地方.

要么分配它,要么使用堆栈上的实例,因为你只是要从它复制.

当然,这整个初始化student2然后s用它覆盖的业务是不必要的复杂,你应该直接修改s.

这个:

student1 = (struct student *)malloc(sizeof(struct student));

在C中写得更好:

student1 = malloc(sizeof *student1);

这将删除无意义(并且可能是危险的)强制转换,并确保大小是该类型的正确大小,将编程器检查的依赖项替换为编译器处理的依赖项.

第三,它是一个典型的"症状"的开始C程序员没有意识到你可以分配结构.所以,而不是

memcpy(s, student2, sizeof *s);

你可以写:

*s = *student2;

让编译器做正确的事情.这可能是性能上的胜利,因为结构可以包含很多填充,分配可以知道并且不能复制,但是memcpy()不能忽略.

Answer 2

复制后将用户名设置为“somebodyelse” 。这仅更改函数“changeme()”内的本地副本。尝试在“changeme()”中打印出student2 ，你就会明白我的意思。