I've always taken this for granted before, but suppose I have:

uint8_t a;
uint8_t b;

if ((a - b) < 0) {
  ...
}

What is the data type of the expression (a - b)? Mr. Godbolt tells me that it's a signed value; is that guaranteed by the any of the C specifications?

AMENDMENT:

I now understand that type promotion will guarantee that (a-b) is an int when a and b are smaller than ints. What if instead a and b are unsigned ints?

unsigned int a;
unsigned int b;

if ((a - b) < 0) {
  ...
}

CodePudding user response：

This expression will have type int, which is signed.

Because both operands have a type smaller than int, both will be promoted to type int, and the result will have type int.

Integer promotions are defined in section 6.3.1.1p2 of the C standard:

The following may be used in an expression wherever an int or unsigned int may be used:

• An object or expression with an integer type (other than int or unsigned int) whose integer conversion rank is less than or equal to the rank of int and unsigned int.
• A bit-field of type _Bool, int, signed int, or unsigned int.

If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions. All other types are unchanged by the integer promotions.

So this means the expression ((a - b) < 0) could potentially evaluate as true.

Had the variables been defined like this:

unsigned int a;
unsigned int b;

Then there would be no promotion and a - b would have unsigned type, meaning ((a - b) < 0) would always be false.