wrap-around functionality in unsigned short

Question

Whenever I run the code :

signed short ssh{-3};
unsigned short ussh{1};

std::cout << ussh + ssh << '\n'; // should be (65535-2+1) = 65534

I get the output as -2 instead of 65534 while the same wrap-around functionality works well with the unsigned long and unsigned int.

Is unsigned short prohibited to perform wrap around or it is just promoted into int??

Answer 1

The rules for integer promotion are very clear for C++. Both values are going to be promoted to a type that has a range that can accept the values without change, which in your case is int.

Nothing is wrapping.

Answer 2

By default, result of arithmetic operation would get promoted to type result can fit in. For bitwise and integer arithmetic operations it's an int or closest "bigger" type which can fit a result without altering it value .

Consider that your code is an equivalent of

{
   auto some_value = ussh + ssh;    // auto should be signed and an int at least
   std::cout << some_value << '\n';
}

Behavior you're looking for can be expressed in this or similar way:

// For complement of 2 it can be considered ssh being 
// cast to unsigned short
ussh += ssh;  
std::cout << ussh << '\n';