【面向对象设计C++--翁恺】22-引用再研究-优快云博客

references as class members

declared without initial value 如果一个类的成员是引用，那么声明的时候没有办法给一个初始值。

must be initialized using constructor initializer list

class X{
public:
    int& m_y;
    X(int& a);
};
X::X(int a):m_y(a){}

returning references

functions can return references

but they better refer to non-local variables!

#include <assert.h>
const int SIZE = 32;
double myarray[SIZE];
double& subscript(const int i){
    return myarray[i];
}

example

main{
    for(int i=0;i<SIZE;i++){
        mayarray[i]=i*0.5;
    }
    double value = subscript(12); 
    subscript(3) = 34.5; // 做左值
}

const in functions arguments

对象送入一个函数，对象可能很大。如果送对象本身，大量数据传递发生，占据很大空间，花费时间做push，某些情况下必须这么做。方法2：送指针进去，函数内部获得对外面对象的控制权，会改变外面的对象。方法3;在指针前面加上const，在函数内无法进行修改该对象，编译器生成的代码不能对对象做修改。方法4：推荐使用，好处是不需要有很多 * 号。常见做法。const 引用。

pass by const value --don’t do it
passing by const reference
```
Person(const string& name,int weight);
```
- don’t change the string object
- more efficient to pass by reference(address) than to pass by value(copy)
- const qualifier protects from change

const reference parameters

可以任意使用那些引用，但是不能做修改，不能做左值。

what if you don’t want the argument changed?

use const modifier

//y is constant!can't be modified 
void func(const int & y,int & z){
    z = z*5;//ok
    y+=5;/error
}

temporary values are const

what you type

void func(int &);
func(i*3);//generates warning or error    i*3 不能做左值

输出9.

what the compiler generates

void func(int &);
const int tem@ = i*3;
func(tem@);//problem--bindling cons ref to non-const argument!

const in function returns

return by const value
- for user defined types,it means “prevent use as an value”
- for built-in’s it means nothing
return by const pointer or reference
- depends on what you want your client to with the return value

C++ 难在内存模型！！！加油掌握它。