C++编译优化备忘

基于GCC测试：http://www.tutorialspoint.com/compile_cpp11_online.php

const A& a=fun() 与 A a= fun()

1、方法fun返回值为 const A& 时，前者比后者 减少一次拷贝构造

 #include <iostream>

 using namespace std;

     class  A
     {

     public:

         A() {
         cout << "--默认构造函数--" << endl;
  }

         A(const A& a) {  cout << "--拷贝构造函数--" << endl; }

         A & operator =(const A &a)
         {
             cout << "--赋值--" << endl;
             return *this;
         }
     };

 class Test
 {
  public:
       const A& GetA(){ cout <<&m_a << endl; return m_a;}

  private:
     A m_a;
 };

 int main()
 {
  Test t;
  A vec;
    cout << "------" << endl;
    A a=t.GetA();
    cout <<&a << endl;
   cout << "------" << endl;
    const A& b=t.GetA();
    cout <<&b << endl;
    return ;
 }

const &

2、方法fun返回值为 A 时，相同。只会生成一个临时变量

 A GetA(){ cout <<&m_a << endl; return m_a;}

　　

 A GetA(){ A a;cout <<&a << endl; return a;}

　　

未实现move构造的类：A a=std::move(fun()) 与  A a=fun()

1、方法fun返回值为 const A& 时，相同，都调用一次拷贝构造

 class Test
 {
  public:
       const A& GetA(){ cout <<&m_a << endl;  return m_a;}

  private:
     A m_a;
 };

 int main()
 {
    Test t;
    cout << "------" << endl;
    A a=t.GetA();
    cout <<&a << endl;
   cout << "------" << endl;
    A b=std::move(t.GetA());
    cout <<&b << endl;
    return ;
 }

move

2、方法fun返回值为 A，前者比后者多一次拷贝构造

 class Test
 {
  public:
       A GetA(){ cout <<&m_a << endl;  return m_a;}

  private:
     A m_a;
 };

 int main()
 {
    Test t;
    cout << "------" << endl;
    A a=t.GetA();
    cout <<&a << endl;
   cout << "------" << endl;
    A b=std::move(t.GetA());
    cout <<&b << endl;
    return ;
 }

move

未实现move构造的类：A a;a=std::move(fun()) 与  A a;a=fun()

与fun的返回值无关，均相同

 class Test
 {
  public:
       A GetA(){ cout <<&m_a << endl;  return m_a;}

  private:
     A m_a;
 };

 int main()
 {
    Test t;
    cout << "------" << endl;
    A a;
    a=t.GetA();
    cout <<&a << endl;
   cout << "------" << endl;
    A b;
    b=std::move(t.GetA());
    cout <<&b << endl;
    return ;
 }

赋值

  const A& GetA(){ cout <<&m_a << endl;  return m_a;}

const A&