运算符号重载问题

[zhao_ivan]

class S
{
public:
S(const int in1):a(in1){}
S operator+ (S& other);
friend const S operator* (S& one , S& theother); //ok
// S operator* (S& one, S& theother); //fail
void display();
private:
int a;
};

S S::operator+(S& other)
{
return (a+other.a);
}

const S operator* (S& one, S& theother) //ok
{
return (one.a * theother.a);
}

//S S::operator* (S& one, S& theother)//fail
//{
// return (one.a * theother.a);
//}

void S::display()
{
cout<<"result is "<< a << endl;
}

void main()
{
S n(5),m(6),o(0);
o = n + m;
o.display();
o = n * m;
o.display();
}

[zhao_ivan]

为什么，注释掉的地方编译无法通过？

[cenlmmx]

S S::operator* ( S& theother)
{
return (this->a * theother.a);
}

[vsfan]

int S::operator* ( S& theother)
{
return (this->a * theother.a);
}

[bugfree]

// S operator* (S& one, S& theother); //fail

this operator return a obj of class S, but you returned a int, one.a*theother.a
so change
return (one.a * theother.a);
to
return S(one.a * theother.a);


/S S::operator* (S& one, S& theother)//fail
//{
// return (one.a * theother.a);
//}

[bugfree]

There is some bad programming habit in your code, so you couse some erro, althrough some of
your code without error, but the programming style is not good, you can not debug your program when to writing a little bigger, I rewrite your code, It is not best, I just want to show you something about programming style:
:-)


=======
#include <iostream>
using namespace std;

class S
{
public:
S(const int in1):a(in1){}
S(const S & other):a(other.a){}

S operator+ (const S& one) { return S(a + one.a);};
S operator* (const S& one) { return S((this->a) * one.a);};
S& operator= (const S& one ){ this->a = one.a; return *this; };
void display();

private:
int a;
};

///////////

void S::display()
{
cout<<"result is "<< a << endl;
}

int main()
{
S n(5),m(6),o(0);
o = n + m;
o.display();
o = n * m;
o.display();
return 0;
}

[zhao_ivan]

感谢各位回答。
我按照上面的办法（bugfree(八个飞飞)）改了还是不行。
编译出错为：binary 'operator *' has too many parameters。
我不理解，‘*’本来就是双目运算符为什么有两个变量不行呢？

还有在运算符号重载时需要注意什么？为什么有的必须是成员变量，有的必须是非静态变量？

[ywchen2000]

int S::operator* (S& one, S& theother)//fail
//{
// return (one.a * theother.a);
//}
or
S S::operator* (S& one, S& theother)//fail
//{
S temp;
// temp.a=one.a * theother.a;
return temp;
//}

[wizlian]

重载后是3元啊~~~~~~

[cenphoenix]

我也想知道，帮你UP一下！

[wizlian]

o = n * m
相当于
o=o.operator*(n,m)

[DanielWYO]

如果在类以外,你的operator完全正确.
但是在类以内,本来二元,加上类的一元,变成三元的了.

[zhao_ivan]

DanielWYO(拼命) 说明白了。

[wizlian]

你还是把他写成friend函数吧~~~~