将指针地址存储在分配的内存中 [英] Store pointer address in malloced memory

问题描述

这感觉像是一个愚蠢的问题，但我无法找到一个干净的解决方案，也无法在大量其他与指针相关的问题中找到类似的问题.

This feels like a silly question, but I just can't work out a clean solution and can't find a similar question in the mass of other pointer related questions.

我有一些未知类型的动态分配的内存，并希望在开始时在其中存储一个指针.malloc 返回的动态内存应该适当对齐，因此我认为在写入分配块的开头时不必担心对齐.

I have some dynamically allocated memory of unknown type and want to store a pointer inside it at the start. Dynamic memory returned by malloc should be suitably aligned so I don't think I have to worry about alignment when writing to the start of the allocated block.

这是我的代码，它可以工作，但我将指针表示为 64 位整数，并希望以更干净/可移植的方式进行:

This is my code, which works, but I'm representing a pointer as a 64-bit integer and want to do it in a more clean/portable way:

void *my_area = malloc(512);
void *my_pointer = get_my_pointer();
((long long *) my_area)[0] = (long long) my_pointer;

推荐答案

转换为 long long 只是额外的负担.改为转换为 void *.

The cast to long long is just extra baggage. Cast to void * instead.

void *my_area = malloc(512);
void *my_pointer = get_my_pointer();
((void **) my_area)[0] = my_pointer;

(我假设这是针对某种自由列表之类的，即您不需要同时使用该结构.)

(I assume that this is for some kind of freelist or the like, i.e., you don't need to use the structure at the same time.)

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