Comparison operation on unsigned and signed integers

Question

See this code snippet

int main() {   unsigned int a = 1000;  int b = -1;  if (a>b) printf("A is BIG! %d\n", a-b);  else printf("a is SMALL! %d\n", a-b);   return 0; }    

This gives the output: a is SMALL: 1001

I don't understand what's happening here. How does the > operator work here? Why is "a" smaller than "b"? If it is indeed smaller, why do i get a positive number (1001) as the difference?

Answer 1

Binary operations between different integral types are performed within a "common" type defined by so called usual arithmetic conversions (see the language specification, 6.3.1.8). In your case the "common" type is unsigned int. This means that int operand (your b) will get converted to unsigned int before the comparison, as well as for the purpose of performing subtraction.

When -1 is converted to unsigned int the result is the maximal possible unsigned int value (same as UINT_MAX). Needless to say, it is going to be greater than your unsigned 1000 value, meaning that a > b is indeed false and a is indeed small compared to (unsigned) b. The if in your code should resolve to else branch, which is what you observed in your experiment.

The same conversion rules apply to subtraction. Your a-b is really interpreted as a - (unsigned) b and the result has type unsigned int. Such value cannot be printed with %d format specifier, since %d only works with signed values. Your attempt to print it with %d results in undefined behavior, so the value that you see printed (even though it has a logical deterministic explanation in practice) is completely meaningless from the point of view of C language.

Edit: Actually, I could be wrong about the undefined behavior part. According to C language specification, the common part of the range of the corresponding signed and unsigned integer type shall have identical representation (implying, according to the footnote 31, "interchangeability as arguments to functions"). So, the result of a - b expression is unsigned 1001 as described above, and unless I'm missing something, it is legal to print this specific unsigned value with %d specifier, since it falls within the positive range of int. Printing (unsigned) INT_MAX + 1 with %d would be undefined, but 1001u is fine.

Answer 2

On a typical implementation where int is 32-bit, -1 when converted to an unsigned int is 4,294,967,295 which is indeed ≥ 1000.

Even if you treat the subtraction in an unsigned world, 1000 - (4,294,967,295) = -4,294,966,295 = 1,001 which is what you get.

That's why gcc will spit a warning when you compare unsigned with signed. (If you don't see a warning, pass the -Wsign-compare flag.)