參考變量和指針問題

Question

我有一個指向一個整數變量的指針。然後我將這個指針分配給一個引用變量。現在當我改變指針指向其他整數變量時，引用變量的值不會改變。誰能解釋爲什麼？

int rats = 101; 
int * pt = &rats; 
int & rodents = *pt;        // outputs  
cout << "rats = " << rats;       // 101 
cout << ", *pt = " << *pt;       // 101 
cout << ", rodents = " << rodents << endl;   // 101 
cout << "rats address = " << &rats;     // 0027f940 
cout << ", rodents address = " << &rodents << endl; // 0027f940 
int bunnies = 50; 
pt = &bunnies; 

cout << "bunnies = " << bunnies;     // 50 
cout << ", rats = " << rats;      // 101 
cout << ", *pt = " << *pt;       // 50 
cout << ", rodents = " << rodents << endl;   // 101 
cout << "bunnies address = " << &bunnies;   // 0027f91c 
cout << ", rodents address = " << &rodents << endl; // 0027f940 

我們將pt分配給兔子，但齧齒動物的值仍然是101.請解釋原因。

Answer 1

線

int & rodents = *pt; 

被創建到什麼pt指向（即rats）的引用。這不是對指針pt的引用。

後來，當您將pt指定爲指向bunnies時，則不會指望rodents引用發生更改。

編輯：爲了說明@Als點，考慮下面的代碼：

int value1 = 10; 
int value2 = 20; 
int& reference = value1; 
cout << reference << endl; // Prints 10 
reference = value2; // Doesn't do what you might think 
cout << reference << endl; // Prints 20 
cout << value1 << endl; // Also prints 20 

第二reference分配確實不變化基準ltself。相反，它將賦值運算符（=）應用於所提到的東西，即value1。

reference將始終指value1並且不能更改。

開始時你的腦袋有點棘手，所以我建議你看看Scott Meyer的優秀書籍Effective C++和More Effective C++。他比我所能解釋的都好得多。