Does Unary + operator do type conversions?

Question

Till now I was believing that there is no use of unary + operator.

But then I came across with following example:

char ch;
short sh;
in         


        

Answer 1

Operator unary + triggers "initial" usual arithmeric conversions, so all integer operands, which type has lower rank than rank of int and unsigned int are promoted to int (or unsigned int if int type does not cover all values of type being promoted on that implementation).