移动或命名返回值优化 (NRVO)?
假设我们有以下代码:
std::vector<int> f()
{
std::vector<int> y;
...
return y;
}
std::vector<int> x = ...
x = f();
这里的编译器似乎有两种方法:
It seems the compiler has two approaches here:
(a) NRVO:破坏 x,然后构造 f() 代替 x.
(b) 移动:在临时空间构造 f(),将 f() 移动到 x,析构 f().
(a) NRVO: Destruct x, then construct f() in place of x.
(b) Move: Construct f() in temp space, move f() into x, destruct f().
根据标准,编译器是否可以自由使用任一方法?
Is the compiler free to use either approach, according to the standard?
推荐答案
编译器可能会将 NRVO 放入临时空间,或者将构造移动到临时空间.从那里它会移动assign x
.
The compiler may NRVO into a temp space, or move construct into a temp space. From there it will move assign x
.
更新:
任何时候您想使用右值引用进行优化,并且您对结果不满意时,请为自己创建一个跟踪其状态的示例类:
Any time you're tempted to optimize with rvalue references, and you're not positive of the results, create yourself an example class that keeps track of its state:
- 构建
- 默认构造
- 搬离
- 销毁
并通过测试运行该类.例如:
And run that class through your test. For example:
#include <iostream>
#include <cassert>
class A
{
int state_;
public:
enum {destructed = -2, moved_from, default_constructed};
A() : state_(default_constructed) {}
A(const A& a) : state_(a.state_) {}
A& operator=(const A& a) {state_ = a.state_; return *this;}
A(A&& a) : state_(a.state_) {a.state_ = moved_from;}
A& operator=(A&& a)
{state_ = a.state_; a.state_ = moved_from; return *this;}
~A() {state_ = destructed;}
explicit A(int s) : state_(s) {assert(state_ > default_constructed);}
friend
std::ostream&
operator<<(std::ostream& os, const A& a)
{
switch (a.state_)
{
case A::destructed:
os << "A is destructed
";
break;
case A::moved_from:
os << "A is moved from
";
break;
case A::default_constructed:
os << "A is default constructed
";
break;
default:
os << "A = " << a.state_ << '
';
break;
}
return os;
}
friend bool operator==(const A& x, const A& y)
{return x.state_ == y.state_;}
friend bool operator<(const A& x, const A& y)
{return x.state_ < y.state_;}
};
A&& f()
{
A y;
return std::move(y);
}
int main()
{
A a = f();
std::cout << a;
}
如果有帮助,请将打印语句放在您感兴趣的特殊成员中(例如复制构造函数、移动构造函数等).
If it helps, put print statements in the special members that you're interested in (e.g. copy constructor, move constructor, etc.).
顺便说一句,如果这对您造成了错误,请不要担心.它对我来说也是段错误.因此,这种特殊设计(返回对局部变量的右值引用)不是一个好的设计.在您的系统上,它可能会打印出A is destructed"而不是段错误.这将是您不想这样做的另一个迹象.
Btw, if this segfaults on you, don't worry. It segfaults for me too. Thus this particular design (returning an rvalue reference to a local variable) is not a good design. On your system, instead of segfaulting, it may print out "A is destructed". This would be another sign that you don't want to do this.
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