為什么復(fù)制構(gòu)造函數(shù)的參數(shù)必須為引用呢，原因如下：
Because if it's not by reference, it's by value. To do that you make a copy, and to do that you call the copy constructor. But to do that, we need to need to make a new value, so we call the copy constructor, and so on...
(You would have infinite recursion because "to make a copy, you need to make a copy".)
來(lái)源：
http://stackoverflow.com/questions/2685854/why-should-the-copy-constructor-accept-its-parameter-by-reference-in-c
老忘記，舉個(gè)例子也許更好理解：
//創(chuàng)建Type類(lèi)型的對(duì)象a
Type a;
//對(duì)a的對(duì)象內(nèi)部數(shù)據(jù)成員賦值
// 調(diào)用復(fù)制構(gòu)造函數(shù)創(chuàng)建對(duì)象b
Type b( a )
由于b的復(fù)制構(gòu)造函數(shù)的參數(shù)被定義為值傳遞，那么首先會(huì)創(chuàng)建一個(gè)Type的臨時(shí)變量，假設(shè)為temp1，然后再傳遞給b的復(fù)制構(gòu)造函數(shù)
Type temp1( a )
由于temp1的復(fù)制構(gòu)造函數(shù)的參數(shù)被定義為值傳遞，那么首先會(huì)創(chuàng)建一個(gè)Type的臨時(shí)變量，假設(shè)為temp2，然后再傳遞給b的復(fù)制構(gòu)造函數(shù)
Type temp2( a )
類(lèi)推,會(huì)出現(xiàn)temp3,temp4,...直至無(wú)窮多！
所以復(fù)制構(gòu)造函數(shù)的參數(shù)必須為引用