使用临时对象复制构造函数需要 [英] Copy Constructor Needed with temp object
问题描述
以下代码仅在复制构造函数可用时工作。
The following code only works when the copy constructor is available.
当我添加打印语句（通过 std :: cout ）并使复制构造函数可用时， （我假设有这样编译器欺骗发生删除不必要的副本）。
When I add print statements (via std::cout) and make the copy constructor available it is not used (I assume there is so compiler trick happening to remove the unnecessary copy).
但是在输出运算符< 和函数 plop 下面（我在其中创建一个临时对象）我看不到需要的复制构造函数。可以有人解释为什么语言需要它，当我通过const引用（或我做错了）一切。
But in both the output operator << and the function plop() below (where I create a temporary object) I don't see the need for the copy constructor. Can somebody explain why the language needs it when I am passing everything by const reference (or what I am doing wrong).
#include <iostream>
class N
{
public:
N(int) {}
private:
N(N const&);
};
std::ostream& operator<<(std::ostream& str,N const& data)
{
return str << "N\n";
}
void plop(std::ostream& str,N const& data)
{
str << "N\n";
}
int main()
{
std::cout << N(1); // Needs copy constructor (line 25)
plop(std::cout,N(1)); // Needs copy constructor
N a(5);
std::cout << a;
plop(std::cout,a);
}
编译器：
[Alpha：〜/ X] myork％g ++ -v
使用内建规格。
目标：i686-apple-darwin10
配置：/ var / tmp / gcc / gcc-5646〜6 / src / configure --disable-checking --enable-werror --prefix = / usr --mandir = / share / man --enable -languages = c，objc，c ++，obj-c ++ --program-transform-name = / ^ [cg] [^ .-] * $ / s / $ / - 4.2 / --with-slibdir = / usr / lib --build = i686-apple-darwin10 --with-gxx-include-dir = / include / c ++ / 4.2.1 --program-prefix = i686-apple-darwin10- --host = x86_64-apple-darwin10 - target = i686-apple-darwin10
线程模型：posix
gcc版本4.2.1（Apple Inc. build 5646）
[Alpha:~/X] myork% g++ -v
Using built-in specs.
Target: i686-apple-darwin10
Configured with: /var/tmp/gcc/gcc-5646~6/src/configure --disable-checking --enable-werror --prefix=/usr --mandir=/share/man --enable-languages=c,objc,c++,obj-c++ --program-transform-name=/^[cg][^.-]*$/s/$/-4.2/ --with-slibdir=/usr/lib --build=i686-apple-darwin10 --with-gxx-include-dir=/include/c++/4.2.1 --program-prefix=i686-apple-darwin10- --host=x86_64-apple-darwin10 --target=i686-apple-darwin10
Thread model: posix
gcc version 4.2.1 (Apple Inc. build 5646)
[Alpha：〜/ X] myork％g ++ t.cpp
t.cpp：在函数'int main（）'中：
t.cpp：10：error： 'N :: N（const N&）'是私人
t.cpp：25：错误：在此上下文中
t.cpp：10：error：'N :: N（const N&）'是私人
t.cpp：26：错误：在此上下文中
[Alpha:~/X] myork% g++ t.cpp
t.cpp: In function ‘int main()’:
t.cpp:10: error: ‘N::N(const N&)’ is private
t.cpp:25: error: within this context
t.cpp:10: error: ‘N::N(const N&)’ is private
t.cpp:26: error: within this context
这是一些实际代码的简化版本。
在实际代码中，我有一个包含std :: auto_ptr的类。这意味着一个复制构造函数接受一个const引用是无效的（没有一些工作），我得到一个错误，指示复制构造函数不可用，因为它：
This is a simplified version of some real code.
In the real code I have a class that contains a std::auto_ptr. This means that a copy constructor that takes a const reference is not valid (without some work) and I was getting an error indicating that the copy constructor was not available because of it:
更改类别：
class N
{
public:
N(int) {}
private:
std::auto_ptr<int> data;
};
错误是：
t.cpp：25：错误：没有匹配函数调用'N :: N（N）'
t.cpp:25: error: no matching function for call to ‘N::N(N)’
推荐答案
从 http://gcc.gnu.org/gcc-3.4 /changes.html
将类型为
的类型的右值绑定到引用时，副本构造函数
的类必须可访问。对于
实例，请考虑以下代码：
When binding an rvalue of class type to a reference, the copy constructor of the class must be accessible. For instance, consider the following code:
class A
{
public:
A();
private:
A(const A&); // private copy ctor
};
A makeA(void);
void foo(const A&);
void bar(void)
{
foo(A()); // error, copy ctor is not accessible
foo(makeA()); // error, copy ctor is not accessible
A a1;
foo(a1); // OK, a1 is a lvalue
}
This might be surprising at first sight, especially since most popular compilers do not correctly implement this rule (further details).
这将在C ++ 1x中修复问题391 。
This will be fixed in C++1x by Core Issue 391.
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