Implicit conversion and operator overload

  • A+

So, I wrote something like this

#include <iostream> using namespace std;  void f(int32_t i) {     cout << "int32: " << i << endl; }  void f(int16_t i) {     cout << "int16: " << i << endl; }  void f(int8_t i) {     cout << "int8: " << i << endl; }  void f(uint32_t i) {     cout << "uint32: " << i << endl; }  void f(uint16_t i) {     cout << "uint16: " << i << endl; }   int main() {     uint8_t i = 0u;     f(i);     return 0; } 

And it printed

int32: 0 

I'm a bit confused:

  • Is this well-defined behaviour, or is it implementation specific?

  • What are the rules that determine which overload is used here and to what type the variable is converted?


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.


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