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
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.