Which integral promotions do take place when printing a char?

Question

I recently read that

unsigned char x=1;
printf(\"%u\",x);

invokes undefined behaviour since due to the format specifier %u, printf expects

Answer 1

Since printf uses a variable argument list, it will be unpacked via va_arg. C++ refers to the C standard for va_arg rules. The C99 Standard says the following:

The va_arg macro expands to an expression that has the specified type and the value of the next argument in the call. The parameter ap shall have been initialized by the va_start or va_copy macro (without an intervening invocation of the va_end macro for the same ap). Each invocation of the va_arg macro modifies ap so that the values of successive arguments are returned in turn. The parameter type shall be a type name specified such that the type of a pointer to an object that has the specified type can be obtained simply by postfixing a * to type. If there is no actual next argument, or if type is not compatible with the type of the actual next argument (as promoted according to the default argument promotions), the behavior is undefined, except for the following cases:

• one type is a signed integer type, the other type is the corresponding unsigned integer type, and the value is representable in both types;
• one type is pointer to void and the other is a pointer to a character type.

Clearly, integer promotions are taken into account when determining whether the actual and expected type match. And signed vs unsigned mismatch is covered by the first bullet point.

Since x = 1 is certainly a value representable by unsigned int, and promotion of unsigned char generates either signed int (if INT_MAX >= UCHAR_MAX) or unsigned int (if INT_MAX < UCHAR_MAX), this is perfectly legal.