当我对结构进行浅表复制时，灵活的数组成员不会被复制

Question

我通过以下方式对我的结构进行了浅拷贝：

struct Student{
        char *name;
        int age;
        Courses *list;  //First course (node)
        Student *friends[];   //Flexible array member stores other student pointers
    }Student;

void shallowCopy(const Student *one){
    Student *oneCopy = malloc(sizeof(one) + 20*sizeof(Student*));

    *oneCopy = *one;     <--------------- ERROR POINTS TO THIS LINE
}

当我检查灵活数组成员 it oneCopy 的第一个元素时，它为空。但是如果我检查原始结构中灵活数组成员的第一个元素，它会成功打印出指针。原始结构的所有其他组件都被复制过来，如名称和链表。只有灵活的数组成员没有被复制。有谁知道我做错了什么？

Answer 1

Does anyone know what I am doing wrong?

正在尝试使用赋值来复制具有灵活数组成员的结构。来自标准（6.7.2.1）：

The assignment *s1 = *s2 only copies the member n [i.e. the part of the struct that isn't a flexible array]; if any of the array elements are within the first sizeof (struct s) bytes of the structure, they might be copied or simply overwritten with indeterminate values.

基本上，当 C 编译器看到一个具有灵活数组成员的结构时，它不知道它到底有多大，所以它认为它足够大以容纳其他成员，加上可能更多：

In particular, the size of the structure is as if the flexible array member were omitted except that it may have more trailing padding than the omission would imply.

这就是 sizeof(*one) 的大小，那是当您执行 *oneCopy = *one;.

时复制的内容的大小

既然你做显然知道整个结构的大小，为了malloc它，只需使用memcpy复制那么多字节。或者，如果您担心这在某种程度上不可移植（老实说我不确定），请执行分配，然后使用循环将每个元素从 one->friends 复制到
oneCopy->friends.

Answer 2

在您的代码片段中，结构声明是错误的。我认为你的意思是结构的 typedef 而不是声明结构的对象。

例如

typedef struct Student
^^^^^^^
{
    char *name;
    int age;
    Courses *list;  //First course (node)
    struct Student *friends[];   //Flexible array memeber stores other student pointers
    ^^^^^^^^^^^^^^
} Student;

这个malloc的调用也是错误的

Student *oneCopy = malloc(sizeof(one) + 20*sizeof(Student*));

应该有

Student *oneCopy = malloc(sizeof( *one ) + 20*sizeof(Student*));
                                  ^^^^^

这是一个演示程序，展示了如何编写该函数

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

typedef struct Student
{
    char *name;
    int age;
//    Courses *list;  //First course (node)
    struct Student *friends[];   //Flexible array memeber stores other student pointers
} Student;

Student * shallowCopy( const Student *one, size_t friends )
{
    Student *oneCopy = malloc( sizeof( Student ) + friends * sizeof( Student * ) );

    *oneCopy = *one;

    memcpy( oneCopy->friends, one->friends, friends * sizeof( Student * ) );

    return oneCopy;
}

int main( void )
{
    Student *one = malloc( sizeof( Student ) + sizeof( Student * ) );

    one->friends[0] = malloc( sizeof( Student ) );

    one->friends[0]->age = 20;

    Student *oneCopy = shallowCopy( one, 1 );

    printf( "Age = %d\n", oneCopy->friends[0]->age );

    free( one->friends[0] );
    free( one );
    free( oneCopy );
}

它的输出是

Age = 20

考虑到该结构最好还包含一个数据成员，用于存储灵活数组中的元素数量。:)

当我对结构进行浅表复制时，灵活的数组成员不会被复制

Flexible array member not getting copied when I make a shallow copy of a struct

c

struct

shallow-copy

flexible-array-member