第40课 前置操作符和后置操作符

1. ++ii++真的有区别吗?

(1)现代编译器会对代码进行优化

  • 对于基础类型前置++和后置++汇编代码几乎是一样的,最终效率完全一样

(2)优化使得最终的二进制程序更加高效

(3)优化后的二进制程序丢失了C/C++的原生语义

(4)不可能从编译后二进制程序还原C/C++程序

 

【编程实验】真的有区别吗?   40-1.cpp

  int i = 0;

013612FB  mov         dword ptr [i],0 

 

    i++;

01361302  mov         eax,dword ptr [i] 

01361305  add         eax,1 

01361308  mov         dword ptr [i],eax 

 

    ++i;

0136130B  mov         ecx,dword ptr [i] 

0136130E  add         ecx,1 

01361311  mov         dword ptr [i],ecx 

 

 

 

2. ++操作符的重载

(1)可利用全局函数成员函数进行重载

(2)重载前置++操作符(如++i不需要额外的参数

(3)重载后置++操作符(如i++需要一个int类型的占位参数(即编译器通过有无这个占位符来区别是重载前置还是后置++操作符)

 

【编程实验】++操作符的重载   40-2.cpp

#include <iostream>

using namespace std;

 

class Test
{

    int mValue;

public:

    Test(int i){mValue = i;}

   

    int value(){return mValue;}

   

    //前置++(如++i),原生语义先自增后取值

    //返回值为引用,无参!

    Test& operator ++() 
    {

        ++mValue;    //先自增

        return *this;//后取值

    }

   

    //后置++(如i++),原生语义先取值后自然

    //返回值为对象,int型参数作占位符,以区别前置和后置++

    Test operator ++(int)
    {

        Test ret(mValue); //先取值

       

        mValue++;         //后自增

       

        return ret;       //注意,这里返回自增之前的对象状态

    }

};

 

int main()
{

    Test t(0);

   

    printf("t.value() = %d\n",(t++).value()); //0;

    //printf("t.value() = %d\n",(++t).value()); //1;

   

    return 0;

}

运行结果:

  

 

3. 真正的区别

(1)对于基础类型的变量

  ①前置++的效率与后置++效率基本相同

  ②根据项目组编码规范进行选择

(2)对于类类型的对象

  ①前置++的效率高于后置++

  ②尽量使用前置++操作符提高程序效率

 

【编程实验】复数类的进一步完善   Complex

//Complex.h

#ifndef _COMPLEX_H_

#define _COMPLEX_H_

 

class Complex
{

private:

    double a;

    double b;

 

public:

    Complex(double a = 0, double b = 0);

    double getA();

    double getB();

    double getModulus();

 

    Complex operator + (const Complex& c);

    Complex operator - (const Complex& c);

    Complex operator * (const Complex& c);

    Complex operator / (const Complex& c);

 

    bool operator == (const Complex& c);

    bool operator != (const Complex& c);

 

    Complex& operator = (const Complex& c);

   

    Complex& operator ++();   //前置++

    Complex operator ++(int); //后置++

};

 

#endif

 

//Complex.cpp

#include "Complex.h"

#include <math.h>

 

Complex::Complex(double a, double b)

{

    this->a = a;

    this->b = b;

}

 

double Complex::getA()

{

    return a;

}

 

double Complex::getB()

{

    return b;

}

 

double Complex::getModulus()

{

    return sqrt(a * a + b * b);

}

   

Complex Complex::operator + (const Complex& c)

{

    double na = a + c.a;

    double nb = b + c.b;

 

    return Complex(na, nb);

}

 

Complex Complex::operator - (const Complex& c)

{

    double na = a - c.a;

    double nb = b - c.b;

 

    return Complex(na, nb);

}

 

Complex Complex::operator * (const Complex& c)

{

    double na = a * c.a - b * c.b;

    double nb = a * c.b - b * c.a;

 

    return Complex(na, nb);

}

 

Complex Complex::operator / (const Complex& c)

{

    double cm = c.a * c.a + c.b * c.b;

    double na = (a * c.a + b * c.b) / cm;

    double nb = (b * c.a - a * c.b) / cm;

 

    return  Complex(na, nb);

}

   

bool Complex::operator == (const Complex& c)

{

    return (a == c.a) && (b = c.b);

}

 

bool Complex::operator != (const Complex& c)

{

    //整个复数对象就两个成员,如果这个2个对象的

    //内存完全相等时,则两个复数相等

    return !(*this == c);

}

 

   

Complex& Complex::operator = (const Complex& c)

{

      if(this != &c)

    {

        a = c.a;

        b = c.b;

    }

    return *this;

}

 

Complex& Complex::operator ++()  //前置++

{

    a = a + 1;

    b = b + 1;

   

    return *this;

}

 

Complex Complex::operator ++(int) //后置++

{

    Complex ret(a, b);

   

    a = a + 1;

    b = b + 1;

   

    return ret;

}

 

//main.cpp

#include <stdio.h>

#include "Complex.h"

 

int main()

{

    Complex c1(0, 0);

    Complex t1 = c1++;

    Complex t2= ++c1;

 

    printf("t1.a = %f, t1.b = %f\n",t1.getA(), t1.getB());//0, 0

    printf("t2.a = %f, t2.b = %f\n",t2.getA(), t2.getB());//2, 2

 

    return 0;

}

运行结果:

  

 

4. 小结

(1)编译优化使得最终的可执行程序更加高效

(2)前置++操作符后置++操作符都可以被重载

(3)++操作符的重载必须符合其原生语义(即前置就先取值再自增,后置应先自增再取值)

  • 对于基础类型前置++后置++效率几乎相同
  • 对于类类型前置++的效率高于后置++
posted @ 2018-12-23 23:36  梦心之魂  阅读(170)  评论(0编辑  收藏  举报