g ++不编译某些嵌套模板

Question

When BREAK is defined, g++ 4.7.2 will not compile the following, which I think is valid C++. It does compile with BREAK defined if the A<U> tmp is changed to something else, like A<int> tmp - while that makes the minimal test case here work, it's no good in my actual application. Is there anything here that is not legal C++?

template <typename T>
class B {

};

template <typename T>
class A {
public:
    template <typename U> B<U> *alloc_B( );
};

template <typename T> template <typename U>
B<U> *A<T>::alloc_B( ) {
    return new B<U>( );
}

#ifdef BREAK
template <typename T>
class C {
public:
    template <typename U> void x(B<U> &b) {
        A<U> tmp;
        B<U> *tmp2;
        tmp2 = tmp.alloc_B<U>( );
        delete tmp2;
    }
};
#endif

int main( ) {
    A<int> a;
    B<float> *bp = a.alloc_B<float>( );
    delete bp;

#ifdef BREAK
    C<int> c;
    B<float> b;

    c.x(b);
#endif
}

Answer 1

The alloc_B function template is a dependent name. You must call it as so:

tmp2 = tmp.template alloc_B<U>( );

That's the problem, and that is why it works when you use A<int> , because the type no longer depends on the template argument U .

Answer 2

This is due to one of C++ 's annoying parsing rules. When it sees tmp.alloc_B<U> , this is not interpreted as a template, but as tmp.alloc_B < U . To fix this, you need to explicitly specific this is a template:

tmp2 = tmp.template alloc_B<U>( );