学习C++笔记246

2.引用更接近const指针，必须在创建时进行初始化，一旦引用和某个变量关联起来，该引用就会一直指向该变量。

int rats =10;

int&rodents = rats;

上面代码实际上是下述代码的伪装表示：

int rats =10;

int*const pr =&rats;

例子：

#include<iostream>

usingnamespace std;

int main()

{

 int rats =100;

 int&rodent = rats;

 cout <<"rats = "<<rats<<", rosent = "<<rodent<<endl;

 cout <<"rats address = "<<&rats<<endl;

 cout <<"rosent address = "<<&rodent<<endl;

 cout <<"==================================="<<endl;

 int bunnies =50;

 rodent = bunnies;

 cout <<"rats = "<<rats<<", rosent = "<<rodent<<", bunnies = "<<bunnies<<endl;

 cout <<"rats address = "<<&rats<<endl;

 cout <<"rosent address = "<<&rodent<<endl;

 cout <<"bunniess address = "<<&bunnies<<endl;

 return0;

}

输出结果：

rats =100, rosent =100

rats address =0xbfce21e4

rosent address =0xbfce21e4

===================================

rats =50, rosent =50, bunnies =50

rats address =0xbfce21e4

rosent address =0xbfce21e4

bunniess address =0xbfce21e8

从结果可以看出，虽然在调用 rodent = bunnies; 后引用 rosent 的值变为 50，但是 rosent 所指向的地址空间还是指向了 rats，没有发生改变，说明 rodent = bunnies; 只是将 bunnies 的值赋值给引用 rodent 所指向的变量，没有改变引用的指向。