我正在研究以下帖子access private member using template trick 。 我想知道应该如何修改代码才能访问多个私有(private)变量 我尝试了以下方法
#pragma once
template<typename Tag, typename Tag::type M>
struct Rob {
friend typename Tag::type get( typename Tag::type) {
return M;
}
};
// use
struct A {
A(int a) :a(a) { }
private:
int a;
int b;
};
// tag used to access A::a
template<typename Tag, typename Member>
struct TagBase {
typedef Member type;
friend type get(Tag);
};
struct A_f : TagBase<A_f, int A::*> { };
template struct Rob<A_f, &A::a>;
template struct Rob<A_f, &A::b>;
int main() {
A a(42);
std::cout << "proof: " << a.*get(A_f()) << std::endl;
}
但我收到以下错误
error C2084: function 'int A::* Rob<A_f,pointer-to-member(0x0)>::get(int A::* )' already has a body
message : see previous definition of 'get'
message : see reference to class template instantiation 'Rob<A_f,pointer-to-member(0x4)>' being compiled
这是link前往演示
最佳答案
这是因为 typename Tag::type
都是 int A::*
,因此两个实例化都定义了相同的函数。
要解决此问题,您需要稍微更改示例,以便它使用多种标记类型:
#include <iostream>
template<typename Tag, typename Tag::type M>
struct Rob {
// Here we receive tag directly
friend typename Tag::type get(Tag) {
return M;
}
};
// use
struct A {
A(int a) :a(a) { }
private:
int a;
int b;
};
// tag used to access A::a
template<typename Tag, typename Member>
struct TagBase {
typedef Member type;
friend type get(Tag);
};
struct A_af : TagBase<A_af, int A::*> { };
struct A_bf : TagBase<A_bf, int A::*> { };
template struct Rob<A_af, &A::a>;
template struct Rob<A_bf, &A::b>;
int main() {
A a(42);
std::cout << "proof: " << a.*get(A_bf()) << a.*get(A_af()) << std::endl;
}
关于C++::使用模板技巧访问多个私有(private)成员,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/65146769/