Implicit conversion and operator overload

The behavior is well-defined, but implementation-specific. With a 16-bit int it would have been different.

The particular rules in the standard are:

[over.best.ics] for overload resolution. [conv.prom] for integral promotion.

In comparing the conversions needed by different overloaded functions, a "promotion" is considered a better conversion sequence than a standard "conversion". Every arithmetic type can promote to at most one other type. (Promotions are also used when passing an argument to a C-style variadic function like printf. The unary + operator can be used to force a promotion of an arithmetic expression, like +n.)

For integer types which are not character types or bool, the promoted type is:

• If int can represent all the values of the original type, then int;
• Otherwise, if unsigned int can represent all the values of the original type, then unsigned int;
• Otherwise, the original type itself (promotion does nothing)

In your example, when comparing the overloaded functions, an "exact match" would be best, but there is no function taking exactly int8_t (or int8_t& or const int8_t&). The promoted type of uint8_t is int, since it's required to support a range much larger than 0-255. And apparently on your system, int32_t is an alias for int, so the function void f(int32_t); requires only a promotion on the argument. The other functions are all viable, but require an integer conversion on the argument. So void f(int32_t); is considered the best overload.

So the technical answer to the question is that it is implementation specific, but only because of the relationship between int and the <cstdint> types, not because of the overload resolution rules.