如何写一个运算符 [英] how to write an operator

问题描述

哪个更好？为什么？


class Complex {

....

friend Complex operator +（const复杂& lhs，const Complex& rhs）;


或


朋友const复杂运算符+（const Complex& lhs，const Complex& rhs ）;


或者是一个而不是一个会员而不是朋友。


Thanx，

Martin < br>

解决方案

Martin Vorbrodt写道：

哪个更好？为什么？

class Complex {
...
朋友复杂算子+（const Complex& lhs，const Complex& rhs）;



朋友const复合体operator +（const Complex& lhs，const Complex& rhs）;
^

这个const没有增加任何价值，有些人声称它会干扰

特定类型的操作。


如果你的意思是复杂的const&'，也不要这样做，因为函数

一般不应该通过引用返回值类型。

或者是其中一个而不是会员而不是朋友。

会员运营商可能无法正常平衡：


5 +复杂（8,2）;


-

Phlip
http://industrialxp.org/community/bi...UserInterfaces

>那个const没有增加任何价值


好​​吧，如果你相信一些人对运营商重载的看法和做为

的整体做"然后它确实使它的行为更像是一个按值返回

内置类型的函数。


int a，b，c;

（a + b）= c; //这不应该编译


复数a，b，c;

（a + b）= c; //如果返回值不是const，这将编译。


由于我从未掌握由值返回的用户定义类型的原因不是

一个左值但它可以成为修改操作的目标。为了使

重载运算符以与内置运算符类似的方式运行，你必须使返回值为常量。


这有价值吗？


我不知道。

" DaKoadMunky" <哒********* @ aol.com>在消息中写道

Phlip

friend const复杂运算符+（const Complex& lhs ，const Complex&

rhs）;

const不添加任何值

好​​吧，如果你相信什么有人说运营商超载，而b $ b就像投注一样。然后它确实使它的行为更像是一个按值返回
内置类型的函数。

int a，b，c;
（a + b）= c; //这不应该编译

复杂的a，b，c;
（a + b）= c; //如果返回值不是const，这将编译。

由于我从未掌握由值返回的用户定义类型的原因是
不是左值，但它可以是修改的目标操作。为了使
重载运算符以与内置运算符
相似的方式运行，必须使返回值为const。

这是否有价值？
<我不知道。

是的，它可以防止意外修改新返回的对象，所以可能

抓住代码编译但可能做错了。


a ++ = b;

which is preferable and WHY?

class Complex {
....
friend Complex operator + (const Complex& lhs, const Complex& rhs);

OR

friend const Complex operator + (const Complex& lhs, const Complex& rhs);

OR either one but a member operator instead of a friend.

Thanx,
Martin

解决方案

Martin Vorbrodt wrote:

which is preferable and WHY?

class Complex {
...
friend Complex operator + (const Complex& lhs, const Complex& rhs);

OR

friend const Complex operator + (const Complex& lhs, const Complex& rhs); ^
That const adds no value, and there are those who claim it interferes with
certain type-specific operations.

If you meant ''Complex const &'', don''t do that either, because functions
generally should not return value-types by reference.
OR either one but a member operator instead of a friend.

A member operator might not balance properly:

5 + Complex(8, 2);

--
Phlip
http://industrialxp.org/community/bi...UserInterfaces

>That const adds no value

Well, if you believe what some people say about operator overloading and "do as
the ints do" then it does make it behave more like a function that returns a
built-in type by value.

int a,b,c;
(a+b) = c; //This should not compile

Complex a,b,c;
(a+b) = c; //This will compile if return value is not const.

For reasons I have never grasped a user-defined type returned by value is not
an lvalue yet it can be the target of modifying operations. To make the
overloaded operator behave in a way analagous to the built-in operators you
must make the return value const.

Does this have value?

I don''t know.

"DaKoadMunky" <da*********@aol.com> wrote in message

Phlip

friend const Complex operator + (const Complex& lhs, const Complex&

rhs);
That const adds no value

Well, if you believe what some people say about operator overloading and "do as the ints do" then it does make it behave more like a function that returns a built-in type by value.

int a,b,c;
(a+b) = c; //This should not compile

Complex a,b,c;
(a+b) = c; //This will compile if return value is not const.

For reasons I have never grasped a user-defined type returned by value is not an lvalue yet it can be the target of modifying operations. To make the
overloaded operator behave in a way analagous to the built-in operators you must make the return value const.

Does this have value?

I don''t know.

Yes, it prevents accidentally modifying a newly returned object, so might
catch code which compiles but may do the wrong thing.

a++ = b;

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