复制构造函数未定义调用

Question

这是完整的示例（源书：C++ From Ground Up，第 3 版，第 219-320 页）：

class sample {
    char *s;
    public:
        sample(); // normal constructor
        sample(const sample &ob); // copy constructor
        ~sample( ) { cout << "s: " << s <<" ,Freeing s\n"; if(s) delete [] s;}
        void show() { cout << s << "\n"; }
        void set(char *str);
        sample operator=(const sample &ob); // overload assignment
};
// Normal constructor.
sample::sample() {
    s = new char('\0'); // s points to a null string.
    cout << "Normal constructor: s: " << strlen(s) << endl;
}
// Copy constructor.
sample::sample(const sample &ob) {
    cout << "Copy constructor: ob.s: "<< ob.s << " ,strlen(ob.s): " <<  strlen(ob.s) << "\n";
    s = new char[strlen(ob.s)+1];
    strcpy(s, ob.s);
}
// Load a string.
void sample::set(char *str) {
    s = new char[strlen(str)+1];
    strcpy(s, str);
}
// Overload assignment operator.
sample sample::operator=(const sample &ob) {
    /* If the target memory is not large enough
    then allocate new memory. */
    cout << "operator= strlen(ob.s): " << strlen(ob.s) << " ,strlen(s): " << strlen(s) << endl;

    if(strlen(ob.s) > strlen(s)) {
        cout << "operator= Larger memory of target object. Deleting current...\n";
        delete [] s;
        s = new char[strlen(ob.s)+1];
    }
    strcpy(s, ob.s);

    return *this;
}
// Return an object of type sample.
sample input() {
    char instr[80];
    static sample str;
    cout << "Enter a string: ";
    cin >> instr;
    str.set(instr);

    return str;
}

int main() {
    sample ob;
    // assign returned object to ob
    ob=input(); // This is now OK
    ob.show();

    return 0;
}

但是我不明白为什么调用了 TWO 次复制构造函数。 （运行代码后输出）

Normal constructor: s: 
Normal constructor: s: 
Enter a string: Hello
Copy constructor: ob.s: Hello ,strlen(ob.s): 5
operator= strlen(ob.s): 5 strlen(s): 0
operator= Larger memory of target object. Deleting current...
Copy constructor: ob.s: Hello ,strlen(ob.s): 5 
s: Hello, Freeing s
s: Hello, Freeing s
Hello
s: Hello, Freeing s
s: Hello, Freeing s

我知道它在 input() 函数返回时被调用，并创建临时对象（通过调用复制构造函数），但我不明白为什么第二次，据我所知（但也许我错了）复制构造函数被 NOT 调用以进行赋值操作，但看起来，尽管如此，当 return *this; 被调用时（当重载运算符返回值时)，复制构造函数被调用了吗？ 我错过了什么？

谢谢

Answer 1

代码真的很糟糕。更不用说其他错误了，这就是为什么复制构造函数被调用两次的原因。

1. 第一次从input()返回对象时调用它，因为它是静态的，按值返回，所以不能应用RVO。

2. 第二次调用是由于operator=() 中的return *this; 而发生的，因为出于某种原因它也按值返回了一个对象：

   sample operator=(const sample &ob);
   // ^^^