#include <string>
#include <vector>
using namespace std;
auto f()
{
vector<string> coll{ "hello" };
//
// Must I use move(coll[0]) ?
//
return coll[0];
}
int main()
{
auto s = f();
DoSomething(s);
}
我知道：如果我只是返回coll,那么coll保证在返回时被移动.
不过,我不确定：coll [0]是否也保证在返回时被移动？
更新：
#include <iostream>
struct A
{
A() { std::cout << "constructed\n"; }
A(const A&) { std::cout << "copy-constructed\n"; }
A(A&&) { std::cout << "move-constructed\n"; }
~A() { std::cout << "destructed\n"; }
};
struct B
{
A a;
};
A f()
{
B b;
return b.a;
}
int main()
{
f();
}
gcc 6.2和clang 3.8输出相同：
constructed
copy-constructed
destructed
destructed
解决方法
In the following copy-initialization contexts,a move operation might@H_403_30@ be used instead of a copy operation:
If the expression in a return statement ([stmt.return]) is a (possibly parenthesized) id-expression that names an object with@H_403_30@ automatic storage duration declared in the body or@H_403_30@ parameter-declaration-clause of the innermost enclosing function or lambda-expression,or
[…]
overload resolution to select the constructor for the copy is first@H_403_30@ performed as if the object were designated by an rvalue. If the first@H_403_30@ overload resolution fails or was not performed,or if the type of the@H_403_30@ first parameter of the selected constructor is not an rvalue reference@H_403_30@ to the object’s type (possibly cv-qualified),overload resolution is@H_403_30@ performed again,considering the object as an lvalue.
b.a和coll [0]都不是id表达式.所以没有隐含的动作.如果你想要一个举动,你必须明确地做.
