Question

我正在做一些实验，看看除了复制省略，RVO，NRVO案件之外的复制时间。

所以我写了一些像这样的代码：

class X {
 public:
  X() { std::cout << "Default constructor" << std::endl; }

  X(const X&) { std::cout << "Copy constructor" << std::endl; }

  X(X&&) { std::cout << "Move constructor" << std::endl; }

  X& operator=(const X) {
    std::cout << "Assignment operator" << std::endl;
    return *this;
  }

  X& operator=(X&&) {
    std::cout << "Move assignment operator" << std::endl;
    return *this;
  }

  ~X() { std::cout << "Destructor" << std::endl; }
};

class Y {
 private:
  X x;

 public:
  const X& getX() const {
    std::cout << "getX" << std::endl;
    return x;
  }
};

int main() {
  Y y;
  std::cout << "assign to ref" << std::endl;
  const X& x1 = y.getX();
  (void)x1;
  std::cout << "assign to const" << std::endl;
  const X x2 = y.getX();
  return 0;
}

我收到以下内容作为输出：

Default constructor
assign to ref
getX
assign to const
getX
Copy constructor
Destructor
Destructor

使用gcc或clang编译-O3并尝试-std = c ++ {11,14,17}时都生成相同的输出。

让我感到惊讶的是，在使用y.getX（）时我没想到要执行任何副本。到一个const变量。我经常使用它来简化我对以下代码中的变量及其成员的访问，但我没有在const引用上执行它，而是我只是使用const希望编译器将其视为重命名。

有谁知道为什么要复制该副本？我想到的唯一原因是它使代码线程安全。如果有多个线程使用对象y，那么我对const的赋值毕竟不是那个const。因为它只会引用对象y中的成员x。其他线程可能会更改。但我不确定这是不是真的意图。

Answer 1

要查看RVO对编译器强制使用NRVO的影响，请在下面的以下修改程序中使用-fno-elide-constructors编译器开关。通过常用选项：

Default constructor 1
assign to ref
getX (with id: 1)
x1 (id:1)
assign to const
getX (with id: 1)
Copy constructor 2
x2 (id:2)
make_X copy
Default constructor 3
make_X (with id: 3)
x3 (id:3)
make_X ref
Default constructor 4
make_X (with id: 4)
x4 (id:4)
Destructor 4
Destructor 3
Destructor 2
Destructor 1

但是使用NRVO你会得到：

Default constructor 1
assign to ref
getX (with id: 1)
x1 (id:1)
assign to const
getX (with id: 1)
Copy constructor 2
x2 (id:2)
additional 1
Default constructor 3
make_X (with id: 3)
Move constructor 4
Destructor 3
Move constructor 5
Destructor 4
x3 (id:5)
additional 2
Default constructor 6
make_X (with id: 6)
Move constructor 7
Destructor 6
x4 (id:7)
Destructor 7
Destructor 5
Destructor 2
Destructor 1

代码示例：

#include <iostream>
int global_id;
class X {
public:
    X() : id(++global_id) {
        std::cout << "Default constructor " << id << std::endl;
    }
    X(const X&) : id(++global_id) {
        std::cout << "Copy constructor " << id << std::endl;
    }
    X(X&&) : id(++global_id) {
        std::cout << "Move constructor " << id << std::endl;
    }
    X& operator=(const X&) {
        std::cout << "Assignment operator " << id << std::endl;
        return *this;
    }
    X& operator=(X&&) {
        std::cout << "Move assignment operator " << id << std::endl;
        return *this;
    }
    ~X() {
        std::cout << "Destructor " << id << std::endl;
    }
    int id;
};

class Y {
    X x;
public:
    const X& getX() const {
        std::cout << "getX (with id: " << x.id << ')' << std::endl;
        return x;
    }
    X make_X() const {
        X extra;
        std::cout << "make_X (with id: " << extra.id << ')' << std::endl;
        return extra;
    }
};

int main()
{
    Y y;
    std::cout << "assign to ref" << std::endl;
    const X& x1 = y.getX();
    std::cout << "x1 (id:" << x1.id << ")\n";
    (void) x1;
    std::cout << "assign to const" << std::endl;
    const X x2 = y.getX();
    std::cout << "x2 (id:" << x2.id << ")\n";
    std::cout << "make_X copy" << std::endl;
    const X x3 = y.make_X();
    std::cout << "x3 (id:" << x3.id << ")\n";
    std::cout << "make_X ref" << std::endl;
    const X& x4 = y.make_X();
    std::cout << "x4 (id:" << x4.id << ")\n";
    return 0;
}

如你所见，RVO真的只能使用局部变量。

成员访问的C ++副本

1 个答案: