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>{};
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.
For an expression
e,the type denoted bydecltype(e)is defined as
follows:
- if
eis 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;- […]
Otherwise,if the expression
std::tuple_size<E>::valueis a
well-formed integral constant expression […] Given the typeTi
designated bystd::tuple_element<i,E>::type,eachviis a
variable of type “reference toTi” 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 isTi.
