当我返回传递给函数的引用参数时，内部会发生什么？

Question

Let's say I have the following function:

int f(int& x)
{
    x++;
    return x;
}

First question, does the return statement copy the value of the reference x since the function is not returning an int&.

Second question, why can't I call the function with a literal? What if I did want to call the function sometimes with a literal sometimes with a value? I tried overloading with

int f(int x)
{
  x++;
  return x;
}

It compiles if I call with literal. But when I pass a variable I get ambiguous call error. Is there a way to accomplish this?

Answer 1

Maybe this is what you want:

int F(const int& x) { return x + 1; }
int F(int& x) { x++; return x; }