Question

Does casting a variable to another type return a temporary copy of that variable? if so then why can’t you reference the temporary variable to a function.

void func(int &i) //error converting parameter 1 from int to int&
{
}

int main()
{
    double d = 6.8;
    func(int(d));
}

Answer 1

Yes casting returns an rvalue (temporary value), but a mutable reference needs an lvalue.

Try this instead:

int main() {
   double d = 6.8;
   {
      int v = d;
      func(v);
      d = v; // if the change needs to be reflected back to d.
      // note that, even if `func` doesn't change `v`, 
      // `d` will always be truncated to 6.
   }
}

If func is not going to modify i, the input argument should be a const reference, which can accept an rvalue.

void func(const int& i);

(but for primitives func(int i) is going to be more efficient.)