c++ - 您自己的类型的结构化绑定(bind)不是结构或元组(通过公共(public)成员函数)

Question

我正在浏览 Herb Sutter 的

旅程:走向更强大、更简单的 C++ 编程

Structure Binding节

为了理解这个概念。最好是写一个我试过但出现一些错误的程序

Just want to try how to use structure binding on class with private data .Please ignore the below example.if any example you can provide

#include<iostream>
#include<string>
using namespace std;

class foobar {
public:
    foobar() { cout << "foobar::foobar()\n"; }
    ~foobar() { cout << "foobar::~foobar()\n"; }

    foobar( const foobar &rhs )
    { cout << "foobar::foobar( const foobar & )\n"; }
    void ival( int nval, string new_string ) { _ival = nval;s=new_string; }

private:
    int _ival;
    string s;
};

foobar f( int val,string new_string ) {
    foobar local;
    local.ival( val,new_string );
    return local;
}

template<> struct tuple_element<0,foobar> { using type = int; };
template<> struct tuple_element<1,foobar> { using type = string; };



 // 2. Now add get<> support (using C++17, because why not; it’s better
 // than =delete’ing the primary get<> template and adding specializations)
 template<int I>
 auto get(const foobar&x) {
 if      constexpr(I == 0) return x._ival;//'_ival' is a private member of 'foobar'
 else if constexpr(I == 1) return x.s;//'s' is a private member of 'foobar'
 }


int main(){
    foobar ml = f( 1024,"hello" );
    auto [ n, s] = f( 1024,"hello" );//Cannot decompose non-public member '_ival' o
    return 0;
}

错误

if constexpr(I == 0) return x._ival;//'_ival' is a private member of 'foobar'

else if constexpr(I == 1) return x.s;//'s' is a private member of 'foobar'

auto [ n, s] = f( 1024,"hello" );//Cannot decompose non-public

需要帮助

1.如果有人能详细说明他在这些方面实际上想做什么(请引用提供的链接)

// 2. Now add get<> support (using C++17, because why not; it’s better
// than =delete’ing the primary get<> template and adding specializations)
template<int I>
auto get(const S&) {
   if      constexpr(I == 0) return x.i;
   else if constexpr(I == 1) return string_view{x.c}; }
   else if constexpr(I == 2) return x.d;
}

2.关于如何修复上述示例的错误的任何建议

Answer 1

这里有很多问题。

首先，为了符合结构化绑定(bind)的条件，您需要特化tuple_size:

namespace std {
    template <> struct tuple_size<foobar> : std::integral_constant<size_t, 2> { };
}

接下来，您对 tuple_element 的特化也必须在 namespace std 中:

namespace std {
    template <> struct tuple_size<foobar> : std::integral_constant<size_t, 2> { };

    template <> struct tuple_element<0,foobar> { using type = int; };
    template <> struct tuple_element<1,foobar> { using type = std::string; };
}

接下来，如果您要像往常一样访问私有(private)成员，您的get 必须声明为friend 函数:

class foobar {
    template <int I> friend auto get(foobar const& );
};

最后，get() 确实有更好的返回引用，否则你的绑定(bind)最终会做出令人惊讶的事情:

template<int I>
auto const& get(const foobar&x) {
    if      constexpr(I == 0) return x._ival;
    else if constexpr(I == 1) return x.s;
}

与其处理 friendship，不如让 get() 成为公共(public)成员，然后编写您需要的三个重载:

class foobar {
public:
    template <size_t I>
    auto& get() & {
        if constexpr (I == 0) return _ival;
        else if constexpr (I == 1) return s;
    }

    template <size_t I>
    auto const& get() const& {
        if constexpr (I == 0) return _ival;
        else if constexpr (I == 1) return s;
    }

    template <size_t I>
    auto&& get() && {
        if constexpr (I == 0) return std::move(_ival);
        else if constexpr (I == 1) return std::move(s);
    }
};

---

此外，ival() 作为函数也没有意义。您的构造函数应该只接受参数。