关于传递临时对象的一个问题

代码：
#include "stdafx.h"
#include
class X {
int i;
public:
X();
int x();
void x(int x);
};

X::X()
{
i = 1;
}

int X::x()
{
return i;
}

void X::x(int x)
{
i = x;
}

X f()
{
return X();
}

void g1(X& x)
{
// cout << " non-const reference:" << x.x() << endl;
}

void g2(const X& x)
{
// cout << " const reference:" << ends << "get x.i:" << x.x() << endl;

// int temp =3;
// cout << "set x.i:" << temp << endl;
// x.x(temp);
// cout << "get x.i:" << x.x() << endl;
}

int main(int argc, char* argv[])
{
g1(f());
g2(f());

return 0;
}
将成元函数改成const的然后调用：
#include "stdafx.h"
#include
class X {
int i;
public:
X();
int x() const ;
void x(int x);
};

X::X()
{
i = 1;
}

int X::x() const
{
return i;
}
void X::x(int x)
{
i = x;
}
X f()
{
return X();
}

void g1(X& x)
{
cout << " non-const reference:" << x.x() << endl;
}

void g2(const X& x)
{
cout << " const reference:" << ends << "get x.i:" << x.x() << endl;

// int temp =3;
// cout << "set x.i:" << temp << endl;
// x.x(temp);
// cout << "get x.i:" << x.x() << endl;
}

int main(int argc, char* argv[])
{
g1(f());
g2(f());

return 0;
}

注：如果声明一个成员函数为const函数，则等于告诉编译器可以为一个const对象调用这个函数。非const对象也可以调用这个函数，为保证函数的常量性，在函数定义中，如果我们改变对象中的任何成员或调用任何非const成元函数，编译器都会报错
即改为如下是不对的：
#include "stdafx.h"
#include <iostream.h>
class X {
    int i;
public:
    X();
     int x() const;
    void x(int x) const ;
};

X::X()
{
    i = 1;
}

  int X::x() const
{
    return i;
}
 void X::x(int x)  const
{
    i = x;
}
X f()
{
    return X();
}

void g1(X& x)
{
    cout << " non-const reference:" << x.x() << endl;
}

void g2(const X& x)
{
    cout << " const reference:" << ends << "get x.i:" << x.x() << endl;
   
    int temp =3;
    cout << "set x.i:" << temp << endl;
//    x.x(temp);
    cout << "get x.i:" << x.x() << endl;
}

    int main(int argc, char* argv[])
{
        g1(f());
        g2(f());

    return 0;
}

原文转自：http://www.ltesting.net