Why is 0x8000000000000000LL considered unsigned long long by gcc?

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I am compiling a piece of code where the value of the 0x8000000000000000LL literal is used to identify an unknown/unsupported value.

The LL suffix indicates that the value should be interpreted as a (signed) long long (int), but gcc (I have tried with 4.8.5 and 4.1.1) says that the value is of type unsigned long long.

I put a sample code here:

#include <stdio.h>

#define UNKNOWN 0x8000000000000000LL

int main(void){
  long long value = 1000;

  if ((unsigned long long) value == UNKNOWN) {
    puts("Yes, they are different!!");
  }

  if (value == (long long) UNKNOWN) {
    puts("Yes, they are different!!");
  }

  if (value == UNKNOWN) {
    puts("Yes, they are different!!");
  }
  return 0;
}

The result of the compilation with this command gcc -Wsign-compare ll.c is this:

ll.c: In function ‘main’:
ll.c:16:13: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
   if (value == UNKNOWN) {
             ^

Why is the 0x8000000000000000LL literal value considered unsigned?

literals gcc c types
1 Answers

Because that's how it works for integer literals that are hexadecimal or octal and don't fit into the integer type suggested by the suffix.

6.4.4.1p5:

The type of an integer constant is the first of the corresponding list in which its value can be represented.

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Notice that unlike decimal integer literals, hexadecimal and octal literals without the u/U suffix can, when searching for a suitable type, flip the type sign before ascending to the next higher-ranking signed integer type.

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