C++:为什么 decltype (*this) 返回引用? [英] C++: Why decltype (*this) returns a reference?

问题描述

template<typename T>
struct foo{
    void f(){
        decltype(*this) a(*this);
        do_some_test(a);
    }
    T data;
};
//compiler won't accept this

在我的解释中，decltype 应该返回 a 类型，以便我们可以在声明中使用它.但是google说在decltype(x)中，如果x是一个左值，它会返回T& where T> 是 x 的类型.

In my interpretation, decltype should return the a type so that we can use it in declaration. But google says that in decltype(x),if x is a lvalue, it will return T& where T is the type of x.

但是他们设计它来返回引用的目的是什么?此外，我应该怎么做才能在模板中创建与 *this 具有相同类型的类的实例?

But what they designed it to return a reference for? Furthermore, what should I do to create a instance of the class that has the same type as *this in a template?

推荐答案

decltype 推导出 表达式 的类型，除非它应用于变量，在这种情况下它推导出该变量的类型:

decltype deduces the type of expression, unless it is applied to a variable, in which case it deduces the type of that variable:

decltype(e)表示的类型定义如下:

——如果 e 是一个无括号的 id 表达式或一个无括号的类成员访问，decltype(e) 是由 e 命名的实体的类型.如果没有这样的实体，或者如果 e 命名了一组重载函数，则程序格式错误；

— if e is an unparenthesized id-expression or an unparenthesized class member access, decltype(e) is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded functions, the program is ill-formed;

——否则，如果 e 是 xvalue，decltype(e) 是 T&&，其中 T 是 e 的类型；

— otherwise, if e is an xvalue, decltype(e) is T&&, where T is the type of e;

——否则，如果 e 是左值，decltype(e) 是 T&，其中 T是e的类型；

— otherwise, if e is an lvalue, decltype(e) is T&, where T is the type of e;

——否则，decltype(e)是e的类型.

§7.1.6.2 [dcl.type.simple]

§7.1.6.2 [dcl.type.simple]

取消引用一个指针会产生一个左值，因此 decltype 会推导出一个对被指点类型的左值引用:

Dereferencing a pointer yields an lvalue, and therefore decltype will deduce an lvalue reference to the type of the pointee:

一元 * 运算符执行间接操作:应用它的表达式应该是指向对象类型的指针或指向函数类型的指针，结果是引用该对象的左值或表达式指向的函数.

The unary * operator performs indirection: the expression to which it is applied shall be a pointer to an object type, or a pointer to a function type and the result is an lvalue referring to the object or function to which the expression points.

§5.3.1 [expr.unary.op]

§5.3.1 [expr.unary.op]

因此decltype(*p)，对于某个指针p，会推导出对被指针类型的左值引用.

Therefore decltype(*p), for some pointer p, deduces an lvalue reference to the type of the pointee.

如果你想从某个指针p中得到指针的类型，你可以使用:

If you wish to get the type of the pointee from some pointer p, you can use:

std::remove_pointer<decltype(p)>::type

或者:

std::remove_reference<decltype(*p)>::type

或者，在您的示例中，您可以简单地说 foo，不需要类型推导.

Or, in your example, you can simply say foo, type deduction is not required.

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