C++ return statement

Question

Why is the output 6 instead of 4?

int f(int a) {
return ++a;
}
int f(unsigned int a) {
return --a;
}
int main() {
cout<<f(5);
}

Answer 1

Integer literals are by default be created as `int` but code parser considers the literal value when choosing type used.

Your f() has 2 variants, with different parameter type.

When you call f(5) an `int` is passed. You can override the literal default typing by adding specifier U after the value e.g. f(5U). That signifies an `unsigned int`. And you'll have the decrementing f() variant be invoked.

Answer 2

Okay so my understanding now is that:

1) In our coding environment, the 5 in f(5) provides the declaration and value (aka definition?) of int a = 5, which is the same as signed int a = 5.

(or is it assumed that int a was already declared beforehand?)

2) An implicit SIGNED (or physically typed SIGNED) variable CANNOT pass through a function with an unsigned condition.

Answer 3

In the first return it will be 6
In the next return it will be 5.
So I think output will be 5.