获取内置和运算符＆（）的＆类型？

Question

Edit: The answer I've marked below isn't 100% correct, that's actually in a comment: using TPtr = decltype(&std::declval<T&>());

---

I'm trying to use std::conditional<> to get the type of &T , where T may have operator&() , or not.

The simple solution is to not even try, instead specify another template argument:

struct Foo
{
    int operator&() { return 314; }
};

struct Bar { };

template <typename T, typename TPtr = T*>
struct A
{
    T& _t;
    A(T& t) : _t(t) {}
    TPtr data() {
        return &_t;
    }
};

where client code is then

Foo foo;
A<Foo, int> aFoo(foo);
int pFooData = aFoo.data();

Bar bar;
A<Bar> aBar(bar);
Bar* pBarData = aBar.data();

But what I'd rather write is something like:

template <typename T>
struct B
{
    using TPtr = typename std::conditional<has_operator_address_of<T>::value, typename std::result_of<decltype(T::operator&)&()>::type, T*>::type;

    T& _t;
    B(T& t) : _t(t) {}
    TPtr data() {
        return &t;
    }
};

where has_operator_address_of is modeled after is_call_possible . Client code is then

B<Foo> bFoo(foo); // no need to second template argument
pFooData = bFoo.data();

but then the class without operator&() fails to compile:

B<Bar> bBar(bar);
pBarData = bBar.data();

It seems that conditional wants to compile both template arguments, so compilation fails when T doesn't have operator&() . I've tried sprinkling std::enable_if around, but to no avail.

Answer 1

You can use decltype and declval , with a little hackery to force an lvalue to be passed to the & operator (which is required):

Start with #include <utility> and define the following helper function:

template <typename T>
T& as_lvalue(T&& val)
{
    return val;
}

Then:

using TPtr = decltype(&as_lvalue(std::declval<T>()));

( live demo )