将指针地址存储在分配的内存中

Question

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.

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.

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;

Answer 1

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;

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

Answer 2

What will be found in my_area[0] is pointer to something, right ?

Then you can allocate my_area to be of type void ** , which represent a pointer to a pointer containing memory area.

void **my_area = malloc(512 * sizeof(*my_area)); // alloc 512 pointer sized memory blocks
void *my_pointer = get_my_pointer();
my_area[0] = my_pointer;

Answer 3

Define a struct with internal array of 8 bytes. Replace all the long long type references with your custom struct. This way you will not depend on platform-specific size of long long. The struct will be 64 bits on all platforms(you can add #pragma pack if you worry about alignment )