C++ 箭头类型产生左值

Question

根据 C++ Primer，C++ 箭头运算符产生一个左值。此外，产生左值的表达式的 decltype 将产生引用类型。那么为什么以下 decltype 不会导致引用类型。

struct MyStruct {
   string name
};
MyStruct s;
s.name = "aname";
MyStruct* p = &s;
decltype (p -> name) str = s.name; //type of str will be string and not &string although p -> name yields an lvalue

Answer 1

来自 cppreference

If the argument is an unparenthesized id-expression or an unparenthesized class member access, then decltype yields the type of the entity named by this expression. If there is no such entity, or if the argument names a set of overloaded functions, the program is ill-formed.

在您的示例中就是这种情况，因此它将返回成员的基础类型，即 std::string。

如果需要，您可以添加括号，以便 decltype 产生引用:

//'str' is a std::string&
decltype((p->name)) str = s.name;