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。那么为什么下面的 decltype not 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.

您的示例就是这种情况，因此它将 return 成员的基础类型，即 std::string。

如果需要，您可以添加括号，以便 decltype 产生参考：

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

Answer 2

来自 decltype（来自 §7.1.6.2/4 [dcl.type.simple]）：

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.

p->name属于上述情况，所以decltype(p->name)的类型是p->name的类型，是std::string而不是std::string&。

另一方面，decltype((p->name)) 是 std::string& 因为 (p->name) 是一个左值表达式。

C++ 箭头类型产生左值

C++ arrow type yields lvalue

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lvalue