CRTP (c++)的各种错误

Various errors with CRTP (C++)

本文关键字：错误 c++ CRTP      更新时间：2023-10-16

我知道我刚才问了一个问题，但我不知道我做错了什么。我只重写了一小部分，找不到任何错误(使用c++函数在父返回子作为参考)

我代码:

#include <iostream>
#include <stdlib.h>
#include <cstring>
using namespace std;
template<class Derived>
class Entity {
    private:
        string _name;
    public:
        const string& name() const;
        Derived& name( const string& );
        Derived* This() { return static_cast<Derived*>(this); }
};
class Client : Entity<Client> {
    private:
        long int _range;
    public:
        const long int& range() const;
        Client& range( const long int& );
};
const string& Entity::name() const {
    return _name;
}
Derived& Entity::name(const string& name) {
    _name = name;
    return *This();
}
const long int& Client::range() const {
    return _range;
}
Client& Client::range( const long int& range ) {
    _range = range;
    return *this;
}
int main() {
    Client ().name("Buck").range(50);
    return 0;
}

结果:

untitled:25: error: ‘template<class Derived> class Entity’ used without template parameters
untitled:25: error: non-member function ‘const std::string& name()’ cannot have cv-qualifier
untitled: In function ‘const std::string& name()’:
untitled:26: error: ‘_name’ was not declared in this scope
untitled: At global scope:
untitled:29: error: expected constructor, destructor, or type conversion before ‘&’ token
untitled: In function ‘int main()’:
untitled:13: error: ‘Derived& Entity<Derived>::name(const std::string&) [with Derived = Client]’ is inaccessible
untitled:44: error: within this context
untitled:44: error: ‘Entity<Client>’ is not an accessible base of ‘Client’

我将非常感谢你的回答(我的无能可能是由于睡眠不足:D)

您需要这样实现您的专用函数:

template<>
const string& Entity<Client>::name() const {
    return _name;
}
template<>
Client& Entity<Client>::name(const string& name) {
    _name = name;
    return *This();
}

并添加公共继承:

class Client : public Entity<Client>

所以你可以访问name()。

如果你想要通用实现:

template<class x>
const string& Entity<x>::name() const {
    return _name;
}
template<class x>
x& Entity<x>::name(const string& name) {
    _name = name;
    return *This();
}

如果您在模板定义之外定义类模板的成员，那么您需要包含模板规范:

template <class Derived>
const string& Entity<Derived>::name() const {
    return _name;
}
template <class Derived>
Derived& Entity<Derived>::name(const string& name) {
    _name = name;
    return *This();
}

您还需要从Entity公开继承:

class Client : public Entity<Client> {
    // stuff
};
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