关于运算附重载，在线等

[danielqinling]

class myString
{
friend int operator=(int & lhs,const myString & rhs)
{
lhs = 1;
};
private:
char * data;
public:
myString & operator = (const char *);
// ostream & operator << (const myString);
// myString();
void show();

};

[qiuafa]

劫分

[njustar]

赋值运算符好像是单目运算符，真是怀疑上面的程序有没有调试过。

[danielqinling]

我的本意就是想重载一个负值运算符，是能够把一个char * 直接负值给整形

[danielqinling]

main.h:4: `int operator=(int&, const myString&)' must be a nonstatic member
function

[liu_feng_fly]

这样改试试
class myString
{
friend int operator=(int & lhs,const myString & rhs);
private:
char * data;
public:
myString & operator = (const char *);
// ostream & operator << (const myString);
// myString();
void show();

};
int operator=(int & lhs,const myString & rhs)
{
lhs = 1;
};

[zhouhu]

ok

[winespirit]

myString & operator = (const char *);
改为
const myString & operator = (const char *);

[noproblem_jyb]

No, no, no.
the following information comes from The C++ Programming Language:

"11.2.2 Predefinied Meanings for Operators[over.predefinied]
Only a few assumptions are make about the meaning of a user-defined operator. In particular operator=, operator[], operator(), and operator-> must be nonstatic member functions; this ensures that their first operands will be lvalues.
..."

remember, "operator=, ..., must be NONSTATIC MEMBER FUNCTIONs;...";

[bht]

ok

[TopCat]

赋值号是不能作为非成员函数重载的。

而且操作符重载不能针对基本类型。你的这个想法是做不到的。有变通的做法，但是有风险，就是定义myString的operator int()，

然后就可以这样用：int i = myString("123");

但是这种做法并不提倡，具体原因可以去看《More Effective C++》的相关条款。

[jinzdi]

同意楼上的观点。
int operator=(int & lhs,const myString & rhs)
这里重载的赋值号不会影响作为左值的int型变量的。
只能通过隐式转换。