为什么我允许复制unique_ptr? [英] Why am I allowed to copy unique_ptr?
问题描述
可能存在重复:
从函数返回unique_ptr
20.7.1.2 [unique。
//禁用左值复制
unique_ptr(const unique_ptr&) =删除;
unique_ptr& operator =(const unique_ptr&)= delete;
那么,为什么下面的代码编译得很好?
#include< memory>
#include< iostream>
std :: unique_ptr< int> bar()
{
std :: unique_ptr< int> p(new int(4));
return p;
int main()
{
auto p = bar();
std :: cout<<< p<< std :: endl;
}
我这样编译它:
g ++ -O3 -Wall -Wextra -pedantic -std = c ++ 0x kel.cpp
编译器:g ++版本4.6.1 20110908(Red Hat 4.6.1-9)
return std :: move(p);
它调用 unique_ptr(unique_ptr&& amp;)$ c $在主函数中,
bar()
会产生一个临时值,它是一个右值,并且也适当地转移到 main
中的 p
。
Possible Duplicate:
Returning unique_ptr from functions
20.7.1.2 [unique.ptr.single] defines copy constructor like this :
// disable copy from lvalue
unique_ptr(const unique_ptr&) = delete;
unique_ptr& operator=(const unique_ptr&) = delete;
So, why the following code compiles fine?
#include <memory>
#include <iostream>
std::unique_ptr< int > bar()
{
std::unique_ptr< int > p( new int(4));
return p;
}
int main()
{
auto p = bar();
std::cout<<*p<<std::endl;
}
I compiled it like this :
g++ -O3 -Wall -Wextra -pedantic -std=c++0x kel.cpp
The compiler : g++ version 4.6.1 20110908 (Red Hat 4.6.1-9)
In the return statement, if you return a local variable, the expression is treated as an rvalue, and thus automatically moved. It is thus similar to:
return std::move(p);
It invokes the unique_ptr(unique_ptr&&)
constructor.
In the main function, bar()
produces a temporary, which is an rvalue, and is also properly moved into the p
in main
.
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