c – 为什么std :: tuple会分解成右值引用

发布时间：2020-12-16 03:33:14 所属栏目：百科来源：网络整理

导读：为什么std :: tuple会分解成右值引用？ #include tupletemplate typename,typename struct same_type;template typename T struct same_typeT,T {};void foo() { std::tuple tuple(1,'a',2.3,true); auto[i,c,d,b] = tuple; same_typedecltype(i),int {}; sa

为什么std :: tuple会分解成右值引用？

#include <tuple>

template <typename,typename> struct same_type;
template <typename T> struct same_type<T,T> {};

void foo() {
  std::tuple tuple(1,'a',2.3,true);
  auto[i,c,d,b] = tuple;
  same_type<decltype(i),int &&>{};
  same_type<decltype(c),char &&>{};
  same_type<decltype(d),double &&>{};
  same_type<decltype(b),bool &&>{};
}

使用gcc trunk编译时没有错误.
我本来期望普通的类型,例如

same_type<decltype(i),int>{};

Live example

解决方法

GCC错误.应用于结构化绑定的decltype返回引用的类型,对于类似于元组的情况,它是std :: tuple_element返回的确切类型.换句话说,语言在这里非常努力地隐藏这些实际上是引用的事实.

[dcl.type.simple]/4：

For an expression e,the type denoted by decltype(e) is defined as
follows:

if e is an unparenthesized id-expression naming a structured binding ([dcl.struct.bind]),decltype(e) is the referenced type as given in the
specification of the structured binding declaration;

[…]

[dcl.struct.bind]/3：

Otherwise,if the expression std::tuple_size<E>::value is a
well-formed integral constant expression […] Given the type Ti
designated by std::tuple_element<i,E>::type,each vi is a
variable of type “reference to Ti” initialized with the initializer,
where the reference is an lvalue reference if the initializer is an
lvalue and an rvalue reference otherwise; the referenced type is Ti.

（编辑：李大同）

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