複製構造函數和返回臨時對象的C++衝突

Question

1 #include <iostream> 
    2 #include <cmath> 
    3 
    4 using namespace std; 
    5 
    6 class complx 
    7 { 
    8 private: 
    9  double real; 
10  double imag; 
11 public: 
12  complx() {} 
13  complx(double r, double i): real(r), imag(i) {} 
14  complx(complx &c) {real = c.real; imag = c.imag;} 
15  complx operator + (const complx &c) const 
16  { 
17   return complx(real + c.real, imag + c.imag); 
18  } 
19 
20  complx & operator=(const complx & c) 
21  { 
22   real = c.real; 
23   imag = c.imag; 
24 
25   return *this; 
26  } 
27 
28  friend ostream& operator << (ostream &os, const complx &c); 
29 
30  double size() const 
31  { 
32   return sqrt(real*real + imag*imag); 
33  } 
34 }; 
35 
36 ostream & operator << (ostream &os, const complx &c) 
37 { 
38  os << "(" << c.real << "," << c.imag << ")"; 
39 
40  return os; 
41 } 
42 
43 const complx & maximum(const complx &c1, const complx &c2) 
44 { 
45  if (c1.size() > c2.size()) 
46  { 
47   return c1; 
48  } 
49  else 
50  { 
51   return c2; 
52  } 
53 } 
54 
55 
56 int main(void) 
57 { 
58  complx c1(10, 30); 
59  complx c2(13, 25); 
60  complx mx = maximum(c1, c2); 
61 
62  cout << c1 << endl; 
63  return 0; 
64 } 

我不知道這段代碼有什麼問題，它無法通過編譯。只有在註解第14行（複製構造函數）或重寫第17行以返回具體對象時，它纔會通過編譯並正確運行。

Answer 1

問題是complx(complx &c)不會綁定到臨時。要解決，你應該重新定義爲：

complx(const complx &c);

一個建議：使用初始化列表，或者更好，轉發其他構造的一個從原始值建構：

complx(complx &c) {real = c.real; imag = c.imag;} 

應：

complx(complx &c): complx(c.real, c.imag) {} 

Answer 2

拷貝構造函數應該採取一個const參數，以確保您不會更改參數的屬性s在構造函數體內，那麼你的構造函數的簽名應該是classname (const classname &obj)