Question

我从rhalbersma那里得到了这个代码，但它在VC 2010中没有编译。我不知道我做错了什么。

template<typename Derived>
struct enable_crtp
{
private:
    // typedefs

    typedef enable_crtp Base;

public:
    // casting "down" the inheritance hierarchy
    Derived const* self() const
    {                   
        return static_cast<Derived const*>(this);
    }

    // write the non-const version in terms of the const version
    // Effective C++ 3rd ed., Item 3 (p. 24-25)
    Derived* self()
    {
        return const_cast<Derived*>(static_cast<Base const*>(this)->self());
    }      

protected:
    // disable deletion of Derived* through Base* 
    // enable deletion of Base* through Derived*
    ~enable_crtp()
    {
        // no-op
    }
};

template<typename FX>
class FooInterface
    :
    private enable_crtp< FX >
{
public:
    // interface
    void foo() { self()->do_foo(); }
};

class FooImpl
    :
    public FooInterface< FooImpl > 
{
private:
    // implementation
    friend class FooInterface< FooImpl > ;
    void do_foo() { std::cout << "Foo\n"; }
};

class AnotherFooImpl
    :
    public FooInterface< AnotherFooImpl > 
{
private:
    // implementation
    friend class FooInterface< AnotherFooImpl >;
    void do_foo() { std::cout << "AnotherFoo\n"; }
};

template<template<typename> class F, int X>
class BarInterface
    :
    private enable_crtp< F<X> >
{
// interface
void bar() { self()->do_bar(); }    
};

template< int X >
class BarImpl
    :
    public BarInterface< BarImpl, X > 
{
private:
    // implementation
    friend class BarInterface< ::BarImpl, X >;
    void do_bar() { std::cout << X << "\n"; }    
};

int main()
{
    FooImpl f1;         
    AnotherFooImpl f2;

    BarImpl< 1 > b1;
    BarImpl< 2 > b2;

    f1.foo();
    f2.foo();
    b1.bar();
    b2.bar();

    return 0;
}

Answer 1

template<template<typename> class F, int X>
class BarInterface
: 
private enable_crtp< F<X> >
{
// interface
void bar() { self()->do_bar(); }    
};

我猜

    template<template<int > class F, int X>
class BarInterface
: 
private enable_crtp< F<X> >
{
// interface
void bar() { self()->do_bar(); }    
};

然后你需要修复关于访问私有成员函数的两个错误

template<typename Derived>
struct enable_crtp
{
private:
    // typedefs

    typedef enable_crtp Base;

public:
    // casting "down" the inheritance hierarchy
    Derived const* self() const
    {                   
        return static_cast<Derived const*>(this);
    }

    // write the non-const version in terms of the const version
    // Effective C++ 3rd ed., Item 3 (p. 24-25)
    Derived* self()
    {
        return const_cast<Derived*>(static_cast<Base const*>(this)->self());
    }      

protected:
    // disable deletion of Derived* through Base* 
    // enable deletion of Base* through Derived*
    ~enable_crtp()
    {
        // no-op
    }
};

template<typename FX>
class FooInterface
    :
    private enable_crtp< FX >
{
public:
    // interface
    void foo() { self()->do_foo(); }
};

class FooImpl
    :
    public FooInterface< FooImpl > 
{
private:
    // implementation
    friend class FooInterface< FooImpl > ;
    void do_foo() { std::cout << "Foo\n"; }
};

class AnotherFooImpl
    :
    public FooInterface< AnotherFooImpl > 
{
private:
    // implementation
    friend class FooInterface< AnotherFooImpl >;
    void do_foo() { std::cout << "AnotherFoo\n"; }
};

template<template<int > class F, int X>
class BarInterface
    :
    private enable_crtp< F<X> >
{
    // interface
public: void bar() { self()->do_bar(); }    
};

template< int X >
class BarImpl
    :
    public BarInterface< BarImpl, X > 
{
private:
    // implementation
    friend class BarInterface< ::BarImpl, X >;
    void do_bar() { std::cout << X << "\n"; }    
};

int main()
{
    FooImpl f1;         
    AnotherFooImpl f2;

    BarImpl< 1 > b1;
    BarImpl< 2 > b2;

    f1.foo();
    f2.foo();
    b1.bar();
    b2.bar();

    return 0;
}

Answer 2

抱歉，我为GCC-4.4重新制作了此代码，无法检查VC 2010。错误：

模板BarInterface中的模板类声明错误 - 将typename替换为int： template<template<int> class F, int X> class BarInterface
将公开设置为foo()和bar()方法： public: void xxx() { self()->do_xxx(); }
在类enable_crtp<FX>和BarInterface中为基类FooInterface设置公开： public enable_crtp< FX >
在self()和foo()方法中添加调用bar()方法的范围说明： void xxx() { enable_crtp<FX>::self()->do_xxx(); }

最后我得到了代码：

template<typename Derived>
struct enable_crtp
{
private:
// typedefs

    typedef enable_crtp Base;

public:
    // casting "down" the inheritance hierarchy
    Derived const* self() const
    {                   
        return static_cast<Derived const*>(this);
    }

    // write the non-const version in terms of the const version
    // Effective C++ 3rd ed., Item 3 (p. 24-25)
    Derived* self()
    {
        return const_cast<Derived*>(static_cast<Base const*>(this)->self());
    }      

protected:
    // disable deletion of Derived* through Base* 
    // enable deletion of Base* through Derived*
    ~enable_crtp()
    {
        // no-op
    }
};

template<typename FX>
class FooInterface
    :
    public enable_crtp< FX >
{
public:
    // interface
    void foo() { enable_crtp<FX>::self()->do_foo(); }
};

class FooImpl
    :
    public FooInterface< FooImpl > 
{
private:
    // implementation
    friend class FooInterface< FooImpl > ;
    void do_foo() { std::cout << "Foo\n"; }
};

class AnotherFooImpl
    :
    public FooInterface< AnotherFooImpl > 
{
private:
    // implementation
    friend class FooInterface< AnotherFooImpl >;
    void do_foo() { std::cout << "AnotherFoo\n"; }
};

template<template<int> class F, int X>
class BarInterface
    :
    public enable_crtp< F<X> >
{
public:
// interface
void bar() { enable_crtp< F<X> >::self()->do_bar(); }    
};

template< int X >
class BarImpl
    :
    public BarInterface< BarImpl, X > 
{
private:
    // implementation
    friend class BarInterface< ::BarImpl, X >;
    void do_bar() const { std::cout << X << "\n"; }    
};

int main()
{
    FooImpl f1;         
    AnotherFooImpl f2;

    BarImpl< 1 > b1;
    BarImpl< 2 > b2;

    f1.foo();
    f2.foo();
    b1.bar();
    b2.bar();

    return 0;
}

此模板代码将无法编译。有任何想法吗？

2 个答案: