编译器优化 [英] compiler optimization

问题描述

所以我有一个问题。 :)
你能告诉我下面代码应该产生的输出吗？

So I have a question for you. :) Can you tell me the output the following code should produce?

#include <iostream>
struct Optimized
{
    Optimized() { std::cout << "ctor" << std::endl; }
    ~Optimized() { std::cout << "dtor" << std::endl; }
    Optimized(const Optimized& copy) { std::cout << "copy ctor" << std::endl; }
    Optimized(Optimized&& move) { std::cout << "move ctor" << std::endl; }
    const Optimized& operator=(const Optimized& rhs) { std::cout << "assignment operator" << std::endl; return *this; }
    Optimized& operator=(Optimized&& lhs) { std::cout << "move assignment operator" << std::endl; return *this; }
};

Optimized TestFunction()
{
    Optimized a;
    Optimized b = a;
    return b;
}

int main(int argc, char* argv[])
{
    Optimized test = TestFunction();
    return 0;
}

我的第一个回答是：

1. ctor


2. 复制ctor


3. 移动ctor


4. dtor


5. dtor


6. dtor

1. ctor
2. copy ctor
3. move ctor
4. dtor
5. dtor
6. dtor

IS为true，但仅在编译器优化处于关闭状态。当优化打开时，输出完全不同。启用优化后，输出为：

and it IS true, but only if compiler optimization is turned off. When optimization is turned ON then the output is entirely different. With optimization turned on, the output is:

1. ctor


2. 复制ctor


3. dtor


4. dtor

1. ctor
2. copy ctor
3. dtor
4. dtor

通过编译器优化，测试变量返回变量。

With compiler optimization, the test variable is the return variable.

我的问题是，什么条件会导致这种方式不能优化？

My question is, what conditions would cause this to not be optimized this way?

我一直被教导，返回一个结构/类，导致额外的复制构造函数可以更好地通过作为参考传递进行优化，但编译器正在为我做这个。

I have always been taught that returning a struct/class which results in extra copy constructors could better be optimized by being passed in as a reference but the compiler is doing that for me. So is return a structure still considered bad form?

推荐答案

这被称为复制Elision

标准特别允许优化，只要它可以复制/移动（即，该方法是声明和可访问）。

The optimization is specifically allowed by the Standard, as long as it would be possible to copy/move (ie, the method is declared and accessible).

在这种情况下，编译器中的实现通常称为返回值优化。有两种变体：

The implementation in a compiler is generally referred to, in this case, as Return Value Optimization. There are two variations:

• RVO ：当您返回一个临时（当您返回一个名称为
的对象时，Named会返回aa+ someString; ）

• 
  

• RVO: when you return a temporary (return "aa" + someString;)
• NRVO: N for Named, when you return an object that has a name

两者都由主要编译器实现，但后者只能在较高的优化级别启动，因为它更难以检测。

Both are implemented by major compilers, but the latter may kick in only at higher optimization levels as it is more difficult to detect.

因此，要回答关于返回结构体的问题：我会推荐它。考虑：

Therefore, to answer your question about returning structs: I would recommend it. Consider:

// Bad
Foo foo;
bar(foo);

-- foo can be modified here


// Good
Foo const foo = bar();

后者不仅更清楚，还允许 const enforcement！

The latter is not only clearer, it also allows const enforcement!

这篇关于编译器优化的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！