Question

I’m using Visual Studio 2008. I have this class:

template <bool T1>
class Foo {
public:
    void doSomething() {}

    Foo<T1>& operator=(int a) {
        doSomething();
        return *this;
    }
};

But I want that the method operator= be hidden (by simply doing: return *this) if the template parameter T1 is false.

I need that for instances of Foo, the lines:

Foo<false> foo;
foo = 20; //this should give a compilation error

So I tried specializing the class definition:

template<>
class Foo<false> {
private:
    Foo<false>& operator=(int a) {
        return *this;
    }
};

However, by doing this I lose the method doSomething() on instances that are Foo<false>, which is not what I need.

I’ve tried removing the operator= with boost::enable_if, like this:

typename boost::enable_if<
    boost::mpl::bool_<T1>
    , Foo<T1>
>::type&
operator=(int a) {
    callProxy();
    return *this;
}

But that makes me unable to have a class like the following:

class Bar {
public:
   Foo<true> assignable;
   Foo<false> unassignable;
};

I’ve also tried putting both methods in Foo and removing them with boost::enable_if and boost::disable_if, like this:

 template <bool T1>
 class Foo {
 public:
    void doSomething() {}

    typename boost::enable_if<
        boost::mpl::bool_<T1>
        , Foo<T1>
    >::type&
    operator=(int a) {
        doSomething();
        return *this;
    }

 private:
    typename boost::disable_if<
        boost::mpl::bool_<T1>
        , Foo<T1>
    >::type&
    operator=(int a) {
        return *this;
    }
 };

Which didn’t work too (I expected that, but it was worth trying).

So, is it possible to get the behaviour I need, and if it is, how could I do it?

Answer 1

You can statically assert the condition:

Foo<T1>& operator=(int a) {
    BOOST_STATIC_ASSERT(T1);
    doSomething();
    return *this;
}