A: Provided that function parameter is "const reference", compiler create temporary variable in following two ways.

a) The actual argument is the correct type, however it isn't Lvalue

double Cube(const double & num)

{

num = num * num * num;

return num;

}

double temp = 2.0;

double value = cube(3.0 + temp); // argument is a expression and not a Lvalue;

b) The actual argument is of the wrong type, however of a type which can be converted to the correct type

long temp = 3L;

double value = cuberoot ( temp); // long to double conversion