在C 11中,方法可以重载,表示方法被调用的对象的表达式是否为左值或右值.如果我从一个通过rvalue调用的方法返回*,我需要明确地从*这个或者不是？
Foo Foo::method() &&
{
return std::move(*this); // Is this move required or not?
}
解决方法
这个类型总是一个左派：
§9.3.2[class.this] p1
In the body of a non-static (9.3) member function,the keyword
thisis a prvalue expression whose value is the address of the object for which the function is called. The type ofthisin a member function of a classXisX*. […]
§5.3.1[expr.unary.op] p1
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.
所以你需要std :: move如果你想调用move构造函数.
#include <iostream>
#include <utility>
struct test{
test(){}
test(test const&){ std::cout << "copy ctor // #1\n"; }
test(test&&){ std::cout << "move ctor // #2\n"; }
test f_no_move() &&{ return *this; }
test f_move() &&{ return std::move(*this); }
};
int main(){
test().f_no_move(); // #1
test().f_move(); // #2
}
使用Clang 3.1(我知道实现ref-qualifiers的唯一编译器),我得到以下输出：
$clang++ -std=c++0x -stdlib=libc++ -pedantic -Wall t.cpp $./a.out copy ctor // #1 move ctor // #2
