Why int plus uint returns uint?

Question

int plus unsigned int returns an unsigned int. Should it be so?

Consider this code:

#include 
#include 

        

Answer 1

If by "should it be" you mean "does my compiler behave according to the standard": yes.

C++2003: Clause 5, paragraph 9:

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows:

• blah
• Otherwise, blah,
• Otherise, blah, ...
• Otherwise, if either operand is unsigned, the other shall be converted to unsigned.

If by "should it be" you mean "would the world be a better place if it didn't": I'm not competent to answer that.