Question

但我无法理解答案！

如果（a-> x）的类型是const double＆amp;为什么这段代码会运行？！

#include <iostream>

struct A { double x; };

int main()
{
    A *a=new A;
    decltype(a->x) x3;
    decltype((a->x)) x4 = x3; // is it really const double& ??
    x4=3;// no error !

    const double& x5=x3;
    x5=5;//error 
}

Answer 1

评论中回答了这个问题。但是，我想教给未来的读者如何自己回答这个问题。在此示例中添加一些代码可以使编译器本身告诉您类型是什么：

#include <type_traits>
#include <typeinfo>
#include <iostream>
#ifndef _MSC_VER
#   include <cxxabi.h>
#endif
#include <memory>
#include <string>
#include <cstdlib>

template <typename T>
std::string
type_name()
{
    typedef typename std::remove_reference<T>::type TR;
    std::unique_ptr<char, void(*)(void*)> own
           (
#ifndef _MSC_VER
                abi::__cxa_demangle(typeid(TR).name(), nullptr,
                                           nullptr, nullptr),
#else
                nullptr,
#endif
                std::free
           );
    std::string r = own != nullptr ? own.get() : typeid(TR).name();
    if (std::is_const<TR>::value)
        r += " const";
    if (std::is_volatile<TR>::value)
        r += " volatile";
    if (std::is_lvalue_reference<T>::value)
        r += "&";
    else if (std::is_rvalue_reference<T>::value)
        r += "&&";
    return r;
}

#include <iostream>

struct A { double x; };

int main()
{
    A *a=new A;
    std::cout << "decltype(a->x) has type " <<  type_name<decltype(a->x)>() << '\n';
    std::cout << "decltype((a->x)) has type " <<  type_name<decltype((a->x))>() << '\n';
//     decltype(a->x) x3;
//     decltype((a->x)) x4 = x3; // is it really const double& ??
//     x4=3;// no error !
// 
//     const double& x5=x3;
//     x5=5;//error 
}

输出：

decltype(a->x) has type double
decltype((a->x)) has type double&

decltype和括号答案是错误的？

1 个答案: