試圖理解移動語意

我正在研究自 C 11 以來引入的“移動語意”。我撰寫了一個示例代碼，用于使用普通建構式、復制建構式和移動建構式創建類物件。

/* Useless.cpp */
#include <iostream>

using namespace std;

// interface
class Useless {
private:
    int n; // number of elements
    char* pc; // pointer to data
    static int ct; // number of objects
    void ShowObject() const;
public:
    Useless();
    explicit Useless(int k);
    Useless(int k, char ch);
    Useless(const Useless & f); // regular copy constructor
    Useless(Useless && f); // move constructor
    ~Useless();

    Useless operator (const Useless & f) const;
    void ShowData() const;
};

int Useless::ct = 0;

Useless::Useless() {
      ct;
    n = 0;
    pc =nullptr;
    cout << "default constructor called; number of objects" << ct << endl;
    ShowObject();
}

Useless::Useless(int k) : n(k) {
      ct;
    pc = new char[n];
    cout << "int constructor called; number of objects: " << ct << endl;
    ShowObject();
}

Useless::Useless(int k, char ch) : n(k) {
      ct;
    pc = new char[n];
    for (int i=0; i< n; i  ) {
        pc[i] =  ch;
    }
    cout <<"int, char constructor called; number of objects: " << ct << endl;
    ShowObject();
}

Useless::Useless(const Useless & f) : n(f.n) {
      ct;
    pc = new char[n];

    for (int i =0; i< n; i  ) {
        pc[i] = f.pc[i]; // deep copy 
    }
    cout << "copy const called; number of objects: " << ct << endl;
    ShowObject();
}

Useless::Useless(Useless && f) : n(f.n) {
      ct;
    pc = f.pc; // steal address (shallow copy)
    f.pc = nullptr;
    f.n = 0;
    cout << "move const called; number of objects: " << ct << endl;
    ShowObject();
}

Useless::~Useless() {
    cout << "destructor called; objects left: " << --ct << endl;
    cout << "deleted object:\n";
    ShowObject();
    delete [] pc;
}

Useless Useless::operator (const Useless & f) const {
    cout << "Entering operator () \n";
    Useless temp = Useless (n f.n);

    for (int i=0; i < n; i  ) {
        temp.pc[i] = pc[i]; // copy the current object's characters
    }

    for (int i=n; i < temp.n; i  ) {
        temp.pc[i] = f.pc[i-n]; // copy the argument object's characters
    }
    cout << "temp object:\n";
    cout <<  "leaving operator ()\n"; 
    return temp;
}

void Useless::ShowObject() const {
    cout << "Number of elements: " << n; 
    cout << "Data address: " << (void *) pc << endl;
}

void Useless::ShowData() const {
    if (n==0) {
        cout << "(object empty)";
    }
    else {
        cout << "data: "; 

        for (int i = 0; i < n; i  ) {
            cout << pc[i] ;
        }
    }
    cout << endl;
}

int main() {
    Useless one (10, 'x');
    Useless two = one; // call the copy constructor
    Useless three(20, 'o'); 
    Useless four(one three); // it should call operator () and then the move constructor

    cout << "object one: ";
    one.ShowData();
    cout << "object two: ";
    two.ShowData();
    cout << "object three: ";
    three.ShowData();
    cout << "object four: ";
    four.ShowData();
    return 0;
}

我預計代碼會生成五個Useless物件（即物件“一”、“二”、“三”、由 operator () 函式創建的臨時物件，以及通過呼叫移動建構式創建的物件“四”）。然而，結果與我的預期不同。實際結果如下：

int, char constructor called; number of objects: 1
Number of elements: 10Data address: 0x56230b195eb0
copy const called; number of objects: 2
Number of elements: 10Data address: 0x56230b1962e0
int, char constructor called; number of objects: 3
Number of elements: 20Data address: 0x56230b196300
Entering operator () 
int constructor called; number of objects: 4
Number of elements: 30Data address: 0x56230b196320
temp object:
**leaving operator ()**
object one: data: xxxxxxxxxx
object two: data: xxxxxxxxxx
object three: data: oooooooooooooooooooo
object four: data: xxxxxxxxxxoooooooooooooooooooo
destructor called; objects left: 3
deleted object:
Number of elements: 30Data address: 0x56230b196320
destructor called; objects left: 2
deleted object:
Number of elements: 20Data address: 0x56230b196300
destructor called; objects left: 1
deleted object:
Number of elements: 10Data address: 0x56230b1962e0
destructor called; objects left: 0
deleted object:
Number of elements: 10Data address: 0x56230b195eb0

我的期望是程式通過呼叫上面定義的移動建構式生成第五個物件（物件“四”）（以“一 三”的結果作為輸入引數）。我將不勝感激任何理解結果的線索。

uj5u.com熱心網友回復：

這就是復制省略和回傳值優化。請參閱復制省略。

在 gcc/clang 中，您可以使用禁用它-fno-elide-constructors，然后您將獲得第 5 個物件。

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