Printing a float's bytes...

Code:

---

#include <stdio.h>

int main() {
  float f = -118.625f;
  char* c = (char*)&f;
  unsigned int i = 0;
  for (i = 0; i < sizeof(float); ++i) {
    printf("%u\n", c[i]);
  }
  getchar();
  return 0;
}

---

1) I made this code based on the article at Wikipedia (http://en.wikipedia.org/wiki/IEEE_754).
2) As far as I know, my machine is little endian.
3) According to 1), the binary representation for -118.625f is 11000010 11101101 01000000 00000000.
4) From 2) and 3), I expected my code above to print the values of those bytes - the first one would be 0 (00000000), the second one would be 64 (01000000), the third one would be 237 (11101101) and the last one would be 194 (11000010).

However, when I compile and execute the code above, it gives me this output:

0
64
4294967277
4294967234

You've probably guessed by now what my question is: why isn't the output the same as I expected it to be?

Thank you in advance.

Try making your "char" into "unsigned char", or perhaps, use a "int *p = (int *)&f" and print it as with "%08x" for a 32-bit hex number.

--
Mats

Oh, but of course! Having an unsigned char* instead of a char* did the trick. So, basically, it was converting a signed type to an unsigned type, which gave me unexpected results.

Thanks for the quick reply.

http://en.wikipedia.org/wiki/Sign_extension

4294967277 = 0xFFFFFFED, 0xED = 237
4294967234 = 0xFFFFFFC2, 0xC2 = 194

gg