negative integer implicitly converted to unsigned type

Question

How would one set/unset enum value for something like the following. Using gcc, I get this annoying warning:

test.c:37: warning: negative integer implicitly converted to unsigned type
test.c:39: warning: negative integer implicitly converted to unsigned type
test.c:41: warning: negative integer implicitly converted to unsigned type
test.c:43: warning: negative integer implicitly converted to unsigned type

code is:

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

typedef enum {
 ONE = 0x1,
 TWO = 0x2,
 THREE = 0x4,
 FOUR = 0x8,
} options;

static const char *byte_to_binary (int x)
{
  int z;
  static char b[9];
  b[0] = '\0';

  for (z = 256; z > 0; z >>= 1)
    {
    strcat(b, ((x & z) == z) ? "1" : "0");
    }

  return b;
}

int main(int argc, char *argv[])
{
  options o = 0;
  printf( "%s\n", byte_to_binary(o));
  o |= ONE;
  printf( "%s\n", byte_to_binary(o));
  o |= TWO;
  printf( "%s\n", byte_to_binary(o));
  o |= THREE;
  printf( "%s\n", byte_to_binary(o));
  o |= FOUR;
  printf( "%s\n", byte_to_binary(o));
  o &= ~FOUR;
  printf( "%s\n", byte_to_binary(o));
  o &= ~THREE;
  printf( "%s\n", byte_to_binary(o));
  o &= ~TWO;
  printf( "%s\n", byte_to_binary(o));
  o &= ~ONE;
  printf( "%s\n", byte_to_binary(o));

  return 0;
}

Answer 1

Since your enum does not contain any negative integer constants, I guess GCC has given unsigned int type to your enums. Now the expressions like

o &= ~FOUR

are equivalent to

o = o & ~FOUR

On the RHS, o is unsigned int and ~FOUR is signed int and by type conversion rules, signed int will be converted to unsigned int. Also ~FOUR is a negative number and hence you get a warning for implicit conversion of a negative number to unsigned type.

If you are sure of your logic, you need not worry about the warnings or you can convert your enum to signed by having a dummy enum which equals to a negative number.

Something like

typedef enum {
 DUMMY =-1,
 ONE = 0x1,
 TWO = 0x2,
 THREE = 0x4,
 FOUR = 0x8,
} options;

Also, your code has run-time buffer overflow problems. In function byte_to_binary you are checking for 9 bits, but your buffer is also 9 bytes. It has to be 10 bytes, one for the terminating null. Make it static char b[10]; and everything works fine