#include <iostream>
using namespace std;
class Test
{
    int mValue;
public:
    Test(int i)
    {
        mValue = i;
    }
    int value()
    {
        return mValue;
    }
    Test& operator ++ ()
    {
        ++mValue;
        return *this;
    }
    Test operator ++ (int)
    {
        Test ret(mValue);
        mValue++;
        return ret;
    }
};
int main()
{
    Test t(0);
    Test m(0);
    Test tt = t++;
    cout << "tt = " << tt.value() << endl;
    cout << "t = " << t.value() << endl;
    Test mm = ++m;
    cout << "mm = " << mm.value() << endl;
    cout << "m = " << m.value() << endl;
    return 0;
}

輸出結(jié)果如下：

前置++的效率高于后置++，因?yàn)榍爸玫?+沒(méi)有生成額外的對(duì)象，意味著不需要過(guò)多的內(nèi)存，也就是不需要在棧上生成對(duì)象。而后置的++需要?jiǎng)?chuàng)建棧空間上的對(duì)象，占用棧空間，并且需要調(diào)用構(gòu)造函數(shù)，返回后需要調(diào)用析構(gòu)函數(shù)。

四、真正的區(qū)別

對(duì)于基礎(chǔ)類型的變量

前置++的效率與后置++的效率基本相同
根據(jù)項(xiàng)目組編碼規(guī)范進(jìn)行選擇

對(duì)于類類型的對(duì)象

前置++的效率高于后置++
盡量使用前置++操作符提高程序效率

前面寫過(guò)的復(fù)數(shù)類可以進(jìn)一步完善了：

Complex.h：

#ifndef _COMPLEX_H_
#define _COMPLEX_H_
class Complex
{
    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 ++ ();
    Complex operator ++ (int);
};
#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;
    Complex ret(na, nb);
    return ret;
}
Complex Complex::operator - (const Complex& c)
{
    double na = a - c.a;
    double nb = b - c.b;
    Complex ret(na, nb);
    return ret;
}
Complex Complex::operator * (const Complex& c)
{
    double na = a * c.a - b * c.b;
    double nb = a * c.b + b * c.a;
    Complex ret(na, nb);
    return ret;
}
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;
    Complex ret(na, nb);
    return ret;
}
bool Complex::operator == (const Complex& c)
{
    return (a == c.a) && (b == c.b);
}
bool Complex::operator != (const Complex& c)
{
    return !(*this == c);
}
Complex& Complex::operator = (const Complex& c)
{
    if( this != &c )
    {
        a = c.a;
        b = c.b;
    }
    return *this;
}
Complex& Complex::operator ++ ()
{
    a = a + 1;
    b = b + 1;
    return *this;
}
Complex Complex::operator ++ (int)
{
    Complex ret(a, b);
    a = a + 1;
    b = b + 1;
    return ret;
}

test.cpp：

#include <iostream>
#include "Complex.h"
using namespace std;
int main()
{
    Complex a(0, 0);
    Complex b(0, 0);
    Complex aa = a++;
    Complex bb = ++b;
    cout << "aa的實(shí)部為: " << aa.getA() << endl;
    cout << "aa的實(shí)部為: " << aa.getB() << endl;
    cout << "bb的實(shí)部為: " << bb.getA() << endl;
    cout << "bb的實(shí)部為: " << bb.getB() << endl;
    return 0;
}

輸出結(jié)果如下：