当具有公共分配运算符的私有副本ctor时出错 [英] Error when have private copy ctor with public assignment operator

问题描述

你们中的一个可以解释为什么下面的代码无法编译吗?

Can one of you explain why the following piece of code does not compile?

#include <iostream>

using namespace std;

class Foo
{
public:
  Foo() { cout << "Foo::Foo()" << endl << endl; }
  Foo& operator=(const Foo&) { cout << "Foo::operator=(const Foo&)" << endl << endl; }
private:
  Foo(const Foo& b) { *this = b; cout << "Foo::Foo(const Foo&)" << endl << endl; }
};

int main()
{
  Foo foo;

  foo = Foo();
}

我收到的错误:

$ g++ -o copy_ctor_assign copy_ctor_assign.cc && ./copy_ctor_assign
copy_ctor_assign.cc: In function 'int main()':
copy_ctor_assign.cc:10: error: 'Foo::Foo(const Foo&)' is private
copy_ctor_assign.cc:17: error: within this context

注意:当我删除 private: 关键字时，代码会编译，但永远不会调用复制ctor.那么，为什么它在私有状态下会出错呢?

Note: when I remove the private: keyword the code compiles but the copy ctor is never called. So why does it err when it's private?

不确定这是否重要，但我正在使用:

Not sure if it's important but I'm using:

$ g++ --version
g++ (GCC) 4.1.2 20080704 (Red Hat 4.1.2-44)
Copyright (C) 2006 Free Software Foundation, Inc.

推荐答案

您正在初始化来自临时目录的引用.
标准状态:
应该使用非引用副本初始化的规则(8.5)初始化临时文件(8.5.3参数5).

You are initializing a reference from temporary.
The standard states:
The temporary should be initialized (8.5.3 par 5)"using the rules for a non-reference copy initialization (8.5)".

已删除临时副本的副本结构(标准允许.12.8par 5).
但是，该标准明确规定了(12.2标准1):
即使避免创建临时对象(12.8)，也必须遵守所有语义限制，就像创建临时对象一样.[示例:即使未调用复制构造函数，所有语义限制(例如可访问性) (第11条)，则应满足.]"

The copy construction is removed for the temporary (permitted by the standard. 12.8 par 5).
However, the standard clearly states (12.2 par 1):
"Even when the creation of the temporary object is avoided (12.8), all the semantic restrictions must be respected as if the temporary object was created. [Example: even if the copy constructor is not called, all the semantic restrictions, such as accessibility (clause 11), shall be satisfied. ]"

(此外，当寻找正确的报价时，发现此

(also, when looking for the right quote, found this duplicate :)

从标准中添加相关位置

adding relevant location from the standard

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