Question

Why wouldn’t the following compile:

void f(int8_t a)
{
}

void f(int16_t a)
{
}

typedef boost::variant<int8_t, int16_t> AttrValue;

int main()
{
    AttrValue a;
    a = int8_t(1);
    f(a);
}

With the compiler error:

error C2665: 'f' : none of the 2 overloads could convert all the argument types
could be 'void f(int8_t)'
or 'void f(int16_t)'

However, this is OK:

std::cout << a; // f(a);

Where is Where is std::ostream &operator<<(std::ostream &, const AttrValue &) defined and why is it defined?

Answer 1

Overload resolution happens at compile time, when your boost::variant instance could contain either type, so the compiler has no way of knowing whether to call void f(int8_t) or void f(int16_t).

std::cout << a works because in either case it’s calling the same function, std::ostream &operator<<(std::ostream &, const AttrValue &), which inside dispatches on the runtime type of the instance.

You need to write a visitor to perform the dispatching:

struct f_visitor: public boost::static_visitor<void>
{
    template<typename T> void operator()(T t) const { return f(t); }
};

boost::apply_visitor(f_visitor(), a);