c++ - 为什么具有非常量复制构造函数的类不被视为可复制构造？

Question

给定

struct Foo 
{
    Foo(Foo&) {} 
};

std::is_copy_constructible<Foo>::value是false

Foo 具有有效的复制构造函数:来自 draft n4659:

15.8.1 Copy/move constructors [class.copy.ctor]
1
A non-template constructor for class X is a copy constructor if its first parameter is of type X& , const X& ,
volatile X& or const volatile X& , and either there are no other parameters or else all other parameters
have default arguments (11.3.6). [Example: X::X(const X&) and X::X(X&,int=1) are copy constructors.

但是is_copy_constructible测试 is_constructible_v<T, const T&> (const) 根据标准。

为什么具有非常量复制构造函数的类不被视为可复制构造？

Answer 1

虽然这确实是一个有效的复制构造函数，is_copy_constructible type-trait 被定义为给出与 is_constructible_v<T, const T&> 相同的结果，因为它旨在对应于 CopyConstructible标准也定义了这个概念。

在[utility.arg.requirements]/1它说

The template definitions in the C++ standard library refer to various named requirements whose details are set out in Tables 20–27. In these tables, T is an object or reference type to be supplied by a C++ program instantiating a template;[...] and v is an lvalue of type (possibly const) T or an rvalue of type const T.

CopyConstructible 概念在 Table 24 中定义作为

Table 24 — CopyConstructible requirements (in addition to MoveConstructible)
Expression     Post-condition
T u = v;           the value of v is unchanged and is equivalent to u
T(v)                 the value of v is unchanged and is equivalent to T(v)

因此，由于您的对象不能从 const Foo 构造左值，这是 CopyConstructible 的要求之一, 它不被认为是这样。