Question

I have this bit of code that is outputting the wrong results.

#include <stdio.h>
#include <string.h>

int main() 
{
  unsigned char bytes[4];
  float flt=0;

  bytes[0]=0xde;
  bytes[1]=0xad;
  bytes[2]=0xbe;
  bytes[3]=0xef;

  memcpy( &flt, bytes, 4);

  printf("bytes 0x%x float %e\n", flt, flt);
  return 0;
}

the output that I get is

bytes 0xc0000000 float -2.000001e+00

I am expecting to get

bytes 0xdeadbeef float -6.2598534e+18

edit #1
as was pointed out the endianness could be different which would result in the following

bytes 0xefbeadde float -1.1802469e+29

what I don’t understand is the cast from float to unsigned int resulting in 0xc0000000 (the float in the same printf statement being -2.0000 I would attribute to compiler optimization)

this was working before on a different computer. It could be an architecture change.

Answer 1

It is not problem of memcpy.

1. float is allways converted to double when passed over ... of printf, so you just can’t get 4 bytes on most of intel architectures.
2. when you expacting 0xdeadbeef in this code, you assume that your architecture is BIG endian. There are many little endian architectures, for example Intel x86.