char and the usual arithmetic conversion rules

Question

I know this question has been asked and seemingly answered a gazillion times over but I can\'t seem to match the answers to my own experience.

The C standard specifi

Answer 1

Here's the relevant part from C99:

6.3.1 Arithmetic operands
6.3.1.1 Boolean, characters, and integers
1 Every integer type has an integer conversion rank defined as follows:
...
2 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 whose integer conversion rank is less than 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, 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.

I agree it's obscure, but this is the closest you can find w.r.t. conversion of the various kinds of char or short or _Bool to int or unsigned int.

From the same source:

5.1.2.3 Program execution
In the abstract machine, all expressions are evaluated as specified by the semantics. An actual implementation need not evaluate part of an expression if it can deduce that its value is not used and that no needed side effects are produced (including any caused by calling a function or accessing a volatile object).
...
10 EXAMPLE 2 In executing the fragment
char c1, c2;
/* ... */
c1 = c1 + c2;
the ‘‘integer promotions’’ require that the abstract machine promote the value of each variable to int size and then add the two ints and truncate the sum. Provided the addition of two chars can be done without overflow, or with overflow wrapping silently to produce the correct result, the actual execution need only produce the same result, possibly omitting the promotions.