unsigned short + int实现的类型是否已定义？

Question

Given this code,

unsigned short us = 0; 
int i = 0;
auto sum = us + i;

is the type of sum implementation-defined? My reading of the C++11 standard says yes:

• 5.7/1 says that the usual arithmetic conversions are applied.
• 4.13/1 bullets 2 and 3 say that the rank of int is greater than the rank of unsigned short.
• 5/9 bullet 5 subbullet 4 says that if int can represent all values in unsigned short, the unsigned short is converted to int, and the type of sum is int.
• 5/9 bullet 5 subbullet 5 says that if int can't represent all the values in unsigned short, both operands are converted to unsigned int, and the type of sum is unsigned int.

If the above analysis is legitimate, it means that using auto to declare variables initialized with arithmetic expressions on built-in types can lead to implementation-defined results. I'm guessing it would surprise a lot of programmers that the type of sum above is not fully determined by the standard.

Is my reasoning legitimate?

Answer 1

The type is implementation-specific, but not implementation-defined. Implementation-defined means that the implementation must document what it does.