引用（References）

引用（References）

引用指一个对象的另一个名字，他的地址和原对象是一样的。引用主要被用来表示函数的参数和返回值，特别是为了运算符的重载。

下面用代码说明引用的基本概念：

void referencesTest(){

int a = 0;

int& b = a;

cout<<&a<<" "<<a<<"/n";

cout<<&b<<" "<<b<<"/n";

b++;

cout<<"/n"<<&a<<" "<<a<<"/n";

cout<<&b<<" "<<b<<"/n";

}

输出如下：

0012FF24 0

0012FF24 1

int a = 12;
int &ra = a;

int &rr1 = ra; // OK!
// int &&rr2 = ra; // error!
cout << rr1 <<endl;
rr1 = 3;
cout << a << " " << ra << " " << rr1 << endl;
//输出结果rr1: 12
// a b c:       3 3 3

// int &*pri; // error! //不允许对指针的引用
// int &ar[3]; // error! //不允许对数组的引用

References can't be const or volatile, because aliases can't be const or volatile, though a reference can refer to an entity that is const or volatile. An attempt to declare a reference const or volatile directly is an error:

int &const cri = a; // should be an error . . . 
const int &rci = a; // OK

Strangely, it's not illegal to apply a const or volatile qualifier to a type name that is of reference type. Rather than cause an error, in this case the qualifier is ignored:

typedef int *PI; 
typedef int &RI;
const PI p = 0; // const pointer
const RI r = a; // just a reference!

There are no null references, and there are no references to void:

C *p = 0; // a null pointer 
C &rC = *p; // undefined behavior
extern void &rv; // error!