使用结构初始化语法上堆结构 [英] Use of struct initialization syntax for on-heap struct

问题描述

我有这种简单的结构我想初始化堆，并返回作为函数的指针。

I have this simple structure I want to initialize on the heap and return as a pointer in a function.

struct entry {
    const char* const key; // We don't want the key modified in any way
    const void* data;      // But the pointer to data can change
    struct entry* next;
};

有一个问题，我不能释放calloc 它和一个初始成员之一，因为键是一个常量指针。我在什么地方找到这个语法如下：

There is one problem, I cannot calloc it and initialize members one by one because key is a const pointer. I found somewhere this syntax that works:

struct entry* entry = calloc(1, sizeof(struct entry));
*entry = (struct entry) { .key = key, .data = data, .next = NULL };

但我不知道是怎么回事吧：它创建一个匿名结构，然后将其复制到 *项生活的地方？是安全使用或者我应该preFER创建本地结构，然后与的memcpy 复制到正确的位置？

But I don't know what is going on with it: does it create an "anonymous" struct that is then copied to the place where *entry lives? Is that safe to use or should I prefer creating a local struct that is then copied with memcpy to the right location?

推荐答案

分配你presented是不正确的，不应该编译。

The assignment you presented is not correct and should not compile.

为const成员初始化分配的结构的正确方法是分配一些内存，创建一个临时结构条目对象，然后用memcpy的对象复制到分配的内存：

The correct way of initializing an allocated struct with const members is to allocate some memory, create a temporary struct entry object, and then use memcpy to copy the object to the allocated memory:

void* mem = malloc( sizeof( struct entry ) );
struct entry temp = { key , data , NULL };
memcpy( mem , &temp , sizeof( temp ) );
struct entry* e = mem;

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