Question

I want to send/receive a memory address via a socket in C.
What I have is the following:

void *ptr = malloc(122); /* So the pointer points to some valid address */
unsigned char *socketData = NULL;

socketData = (unsigned char*)malloc(sizeof(void*));
memset(socketData, 0, sizeof(void*));

/* ??? How to get the memory address - where ptr points to - to socketData ??? */

I know that the way to print pointer addresses using printf is to use %p, i.e.

printf("%p", ptr);

But this prints for example 0x0021ef1a.
What I want is just the following: 0021ef1a

And on the receiver side: how to transform the received bytes back to a void*?

Ah: and the code should work for 32bit as well for 64bit systems 😉
Furthermore the code should compile using -Wall -Werror…. huh

Thanks for helping!
Have a nice weekend,
jonas

Answer 1

To answer your question, you can convert the raw data back to a pointer simply by copying again:

void *ptr = malloc(42);
void *ptr2 = NULL;

unsigned char *data = malloc(sizeof(void *));

memcpy(data, &ptr, sizeof(void *));

...

memcpy(&ptr2, data, sizeof(void *));

printf("ptr  = %p\n", ptr);
printf("ptr2 = %p\n", ptr2);

Note that in most situations, sending a pointer over sockets makes little sense. The pointer value will be useless to the receiver, unless it happens to be the very same process as the sender. Each process will have a separate virtual address space.

Given your proposed application, I would suggest that on the receiver side, you should consider representing this with a pointer at all, as it will be of no use in that form. Instead, why not store it in a suitably-large integer type?