C++ Printer Plus学习笔记第八章内联函数引用

雨点打透心脏的1/2处 2022-02-02 23:59 246阅读 0赞

示例：

#include <iostream>
    
    // 关键字 inline 代码块必须简短
    inline double square(double x) {return x * x; }
    
    int main()
    {
      using namespace std;
      double a, b;
      double c = 13.0;
    
      a = square(5.0);
      b = square(4.5 + 7.5);
      cout << "a = " << a << ", b = " << b << "\n";
      cout << "c = " << c;
      cout << ", c squared = " << square(c++) << "\n";
      cout << "Now c = " << c << "\n";
      return 0;
    }

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引用

#include <iostream>
    int main()
    {
      using namespace std;
      int rats = 101;
      int& rodents = rats;
      cout << "rats = " << rats;
      cout << ", rodents = " << rodents << endl;
      rodents++;
      cout << "rats = " << rats;
      cout << ", rodents = " << rodents << endl;
    
      cout << "rats address = " << &rats;
      cout << ", rodents address = " << &rodents << endl;
      return 0;
    }

其实引用是指针的伪装：

int rats = 101;

int& rodents = rats;

int\* const prats = &rats  要留意 const是在里面 也就是说 该指针变量不能指向其他地址

parts = rodents

引用变量无法通过赋值语句改变其内存地址 且指针不允许赋值给引用变量。

C++标准中  对于函数形参使用const 修饰符的引用变量  允许传递类型不正确或不是左值的数据。因为C++会创建临时变量

如果没有const修饰符的话 则不允许这么干（有一些旧的编译器只是报错 但是依然通过编译）

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four为结构类型的变量

accumulate返回dup（与four是同一种结构数据类型）的引用

所以如下的赋值语句的顺序是：

accumulate(dup, five) = four;

dup = accumulate(dup, five)

dup = four

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