Quick Q: Implicit conversion and operator overload
Quick A: it tries to convert to int.
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Implicit conversion and operator overload
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.
ACCU’s Overload journal of February 2019 is out. It contains the following C++ related articles.