How to get float address for use with memcpy?

Question

I'm trying to fill byte/char array with memory address of float variable (array length is 4 bytes = pointer), but whatever I do it keeps getting float value instead of address:

float f = 20.0f;

memcpy(data, &f, sizeof(data));

Debugging it with:

printf("Array: %#X, %#X, %#X, %#X", data[0], data[1], data[2], data[3]);

...gives float value (20.0) in hex format:

Array: 0, 0, 0XA0, 0X41

What I need is memory address of float. I tried casting/dereferencing it in some different ways, but can't get it to work...

Answer 1

It's how memcpy works: it takes a pointer to data it will copy. Your data is pointer to float, so you need to pass pointer to pointer to float:

#include <cstring>

int main() {
   float f = 20.0f;
   float* pf = &f;
   char data[sizeof(pf)];
   memcpy(data, &pf, sizeof(data));
}

Answer 2

For a C solution, save the hex data to a compound literal .

#include <stdio.h>

int main(void) {
  char data[sizeof (float*)];
  printf("%p\n", (void*) data);
  float f;
  printf("%p\n", (void*) &f);

  memcpy(data, &(float *){&f}, sizeof (float*));
  for (unsigned i = 0; i<sizeof data; i++) {
    printf("%02hhX ", data[i]);
  }
  puts("");
}

Output

0xffffcba8
0xffffcba4
A4 CB FF FF 00 00 00 00 

The address is valid until the end of the block.
Adjust endian/print order as desired.