Question

我确信这已经得到了解答，但我不是程序员，无法找到/理解正确的答案。假设我有一个庞大的类big，想要重载一个二元运算符，比如operator +。

是否有一种理智的方式让big X=Y+Z直接在X中构建总和，而不是创建临时对象，将其复制到X然后销毁临时对象？
我唯一想到的就是将big包装到另一个类small中，该类将包含指向big的指针并添加int use;个引用数大，所以big对象在use==0时被销毁。并添加另一个赋值运算符，比如说<=进行实际复制。我试图实现它（下面）。它似乎有效，但是我没有经验，我很难预见，会出现什么问题。不应该有更简单的解决方案吗？

代码：

#include <iostream>

// print and execute cmd
#define Do(cmd) cout << "\n\n\t"<< ++line << ".\t" << #cmd << ";\n" << endl; cmd;

// print small object: name(small.id[big.id,u=big.use,x=big.x])
#define Show(avar) cout << #avar << "(" << (avar).id << "[" << ((avar).data==NULL?0:(avar).data->id) << ",u=" << ((avar).data==NULL?0:(avar).data->use) << ",x=" << ((avar).data==NULL?0:(avar).data->x) << "])" 

using namespace std;

class big{
public:
  static int N;   // biggest id in use
  int id;         // unique id for each object
  int use;        // nuber of references to this object
  int x;          // data
  big() __attribute__((noinline))
  {
    id=++N;
    use=1;
    x=0;
    cout << "big.constructor.def: [" << id << ",u=" << use << ",x="<<x<<"]" << endl;
  }
  big(const int& y) __attribute__((noinline))
  {
    id=++N;
    x=y;
    use=1;
    cout << "big.constructor.int: [" << id << ",u=" << use << ",x="<<x<<"]" << endl;
  }
  big(const big& b) __attribute__((noinline))
  {
    id=++N;
    use=1;
    x=b.x;
    cout << "big.constructor.copy: [" << id << ",u=" << use << ",x="<<x<<"]" << endl;
  }
  ~big() __attribute__((noinline))
  {
    if(use>0) throw 99; // destroing referenced data!
    cout << "big.destructor: [" << id << ",u=" << use << ",x="<<x<<"]" << endl;
  }
  friend class small;
};

class small{
public:
  static int N;      // biggest id in use
  int id;            // unique id
  big * data;        // reference to the actual data
  small() __attribute__((noinline))
  {
    id=++N;        
    data=NULL;       // contains no data
    cout << "small.constructor.def: ";
    Show(*this)<< endl;
  }
  small(const int& y) __attribute__((noinline))
  {
    id=++N;
    data=new big (y);  // relies on the big constructor
    cout << "small.constructor.int: ";
    Show(*this)<<endl;
  }
  small(const small& y) __attribute__((noinline))
  {
    id=++N;
    data=y.data;      // new object refers to the same data!!
    if(data!=NULL) 
      ++(data->use);  // new reference added;
    cout << "small.constructor.copy: "; 
    Show(y) << "-->";
    Show(*this) << endl;
  }
  ~small(){
    cout << "small.destructor: ";
    Show(*this)<< endl;
    if(data!=NULL){       // is there data?
      --(data->use);      // one reference is destroyed
      if(data->use == 0)  // no references left, kill the data
    delete data;
    }
  }
  const small& operator= (const small& b) __attribute__((noinline))
  {
    cout << "equal: ";
    Show(*this) << " = ";
    Show(b)<<endl;
    if(data != NULL){     // is there data in the target?
      --(data->use);      // one reference is destroyed
      if(data->use == 0)  // no references left, 
    delete data;      // kill the data
    }
    data=b.data;          // target referenses the same data as the source!
    if(data!=NULL) 
      ++(data->use);      // new references added
    cout << "Done equal: "<<endl;    
    return *this;
  }
  // <= will be used for actual copying the data
  const small& operator<= (const small& b) __attribute__((noinline))
  {
    cout << "Copy: ";
    Show(*this) << " <= ";
    Show(b)<<endl;
    if(data != NULL){     // is there data in the target?
      --(data->use);      // one reference is destroyed
      if(data->use == 0)  // no references left, 
    delete data;      // kill the data
    }
    if(b.data==NULL)     // source has no data
      data=NULL;
    else
      data = new big(*(b.data)); // make new copy of the data 
                                 // via big's copy constructor
    cout << "Done copy: "<<endl;    
    return *this;
  }
  small operator+ (const small& b) __attribute__((noinline))
  {
    cout << "Plus: "; 
    Show(*this) << " + ";
    Show(b)<< endl;
    if(this->data == NULL | b.data == NULL) throw 99; // missing data for +
    small ret(data->x);
    ret.data->x += b.data->x;
    cout << "Return: "; Show(ret)<<endl;
    return ret;
  }
};


int big::N=0;
int small::N=0;

main(){
  int line=0;

  Do(small X(5); small Y(6); small Z(7); small W(X));
  Show(X) << endl;
  Show(Y) << endl;
  Show(Z) << endl;
  Show(W) << endl;

  Do(X=Y; Z<=Y);
  Show(X)<<endl;  
  Show(Y)<<endl;  // X and Y refer to the same data
  Show(Z)<<endl;  // Z has a copy of data in Y

  Do(X=Z; Y=Z);
  Show(X)<<endl;
  Show(Y)<<endl;
  Show(Z)<<endl;  // data previosly in X,Y destroyed

  Do(small* U=new small (17); small* T=new small (*U));
  Show(*U) << endl;
  Show(*T) << endl; // U and T refer to the same big

  Do(delete U);
  Show(*T) << endl; // big stays since there is another reference to it

  Do(delete T);     // big destroyed

  Do(X=(Y+Z)+W);
  Show(X)<<endl;
  Show(Y)<<endl;
  Show(Z)<<endl;  // no extra copying of data occures

  cout << "\n\tEND\n" << endl;
}

输出：

1.  small X(5); small Y(6); small Z(7); small W(X);

big.constructor.int: [1,u=1,x=5]
small.constructor.int: *this(1[1,u=1,x=5])
big.constructor.int: [2,u=1,x=6]
small.constructor.int: *this(2[2,u=1,x=6])
big.constructor.int: [3,u=1,x=7]
small.constructor.int: *this(3[3,u=1,x=7])
small.constructor.copy: y(1[1,u=2,x=5])-->*this(4[1,u=2,x=5])
X(1[1,u=2,x=5])
Y(2[2,u=1,x=6])
Z(3[3,u=1,x=7])
W(4[1,u=2,x=5])


    2.  X=Y; Z<=Y;

equal: *this(1[1,u=2,x=5]) = b(2[2,u=1,x=6])
Done equal: 
Copy: *this(3[3,u=1,x=7]) <= b(2[2,u=2,x=6])
big.destructor: [3,u=0,x=7]
big.constructor.copy: [4,u=1,x=6]
Done copy: 
X(1[2,u=2,x=6])
Y(2[2,u=2,x=6])
Z(3[4,u=1,x=6])


    3.  X=Z; Y=Z;

equal: *this(1[2,u=2,x=6]) = b(3[4,u=1,x=6])
Done equal: 
equal: *this(2[2,u=1,x=6]) = b(3[4,u=2,x=6])
big.destructor: [2,u=0,x=6]
Done equal: 
X(1[4,u=3,x=6])
Y(2[4,u=3,x=6])
Z(3[4,u=3,x=6])


    4.  small* U=new small (17); small* T=new small (*U);

big.constructor.int: [5,u=1,x=17]
small.constructor.int: *this(5[5,u=1,x=17])
small.constructor.copy: y(5[5,u=2,x=17])-->*this(6[5,u=2,x=17])
*U(5[5,u=2,x=17])
*T(6[5,u=2,x=17])


    5.  delete U;

small.destructor: *this(5[5,u=2,x=17])
*T(6[5,u=1,x=17])


    6.  delete T;

small.destructor: *this(6[5,u=1,x=17])
big.destructor: [5,u=0,x=17]


    7.  X=(Y+Z)+W;

Plus: *this(2[4,u=3,x=6]) + b(3[4,u=3,x=6])
big.constructor.int: [6,u=1,x=6]
small.constructor.int: *this(7[6,u=1,x=6])
Return: ret(7[6,u=1,x=12])
Plus: *this(7[6,u=1,x=12]) + b(4[1,u=1,x=5])
big.constructor.int: [7,u=1,x=12]
small.constructor.int: *this(8[7,u=1,x=12])
Return: ret(8[7,u=1,x=17])
equal: *this(1[4,u=3,x=6]) = b(8[7,u=1,x=17])
Done equal: 
small.destructor: *this(8[7,u=2,x=17])
small.destructor: *this(7[6,u=1,x=12])
big.destructor: [6,u=0,x=12]
X(1[7,u=1,x=17])
Y(2[4,u=2,x=6])
Z(3[4,u=2,x=6])

    END

small.destructor: *this(4[1,u=1,x=5])
big.destructor: [1,u=0,x=5]
small.destructor: *this(3[4,u=2,x=6])
small.destructor: *this(2[4,u=1,x=6])
big.destructor: [4,u=0,x=6]
small.destructor: *this(1[7,u=1,x=17])
big.destructor: [7,u=0,x=17]

Answer 1

有，它被称为copy elision。与此案例特别相关的是return value optimization (RVO) named and return value optimization (NRVO)。这意味着在某些情况下返回值时，允许编译器忽略副本。实施一个天真的加法运算符可能会导致RVO。

请注意，这是编译器允许执行的优化，但不保证会发生。但是C ++具有移动语义，它提供了一种形式的方法，通过这种方式，一个（通常是临时的）对象的底层数据可以“移动”到另一个对象，而不会产生不必要的副本。有一篇关于移动语义的文章here。

Answer 2

如果和是一个复合值，另一种方法是让big::operator+返回一个sumOfBig类的实例，它保留指向Y和Z的指针或引用。

sumOfBig可能包含在需要时动态计算总和成分的成员函数。

Answer 3

考虑在这种情况下定义和使用+=：

Big a, b, c;

而不是：

a = b + c;

执行：

a=b;
a+=c;

+=的示例定义：

Big& Big::operator += (const Big& other)
{
   this->a += other.a;
   // ...
   return *this;
}

您可以根据operator +制作operator +=，使其在逻辑上相同。

Answer 4

首先，启用优化并查看您的编译器实际的作用：根据juanchopanza的回答，您可以免费获得（N）RVO。

如果您的编译器无法做到这一点，根据Piotr的回答显式删除中间副本可能会改善问题。

如果你真的需要推迟对任意复杂算术表达式的评估，你需要表达式模板（正如Nicola和我在评论中提到的）：如果你有可变的源对象，你可能需要表达式模板和写入时复制。这些都不是微不足道的。如果你真的需要它们，并且真的找不到已经完成你所需要的库...那么，一旦你研究了它们并开始了，我会留意你的表达模板实现的问题。 / p>

复制大对象，c ++

4 个答案: