赋值运算符问题 [英] assignment operator question
问题描述
当编写Apple类并且赋值运算符不是明确声明的
时，运算符
Apple& operator =（const Apple&）
由编译器创建。这和它之间有什么区别
const Apple& operator =（const Apple&）
除了第二种形式意味着你不能这样做：
Apple a，b ，c;
//假设有一个功能''Apple运营商+（const Apple&，const
Apple&）''
（a + b）= c; //我认为不允许使用第二个赋值运算符。
谢谢，
matthew
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Hi,
When a class Apple is written and the assignment operator is not
explicitly declared, the operator
Apple& operator=(const Apple&)
is created by the compiler. Is there any difference between this and
const Apple& operator=(const Apple&)
other than the second form means that you can''t do this:
Apple a, b, c;
//Assume there is a function ''Apple operator+(const Apple&, const
Apple&)''
(a + b) = c; //not allowed with second assignment operator, I think.
thanks,
matthew
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推荐答案
" Matthew Polder" <毫安*************** @ kodak.com>在留言中写道
新闻：40 *************** @ kodak.com ...
"Matthew Polder" <ma***************@kodak.com> wrote in message
news:40***************@kodak.com...
当编写Apple类并且未明确声明赋值运算符时，运算符
Apple& operator =（const Apple&）
由编译器创建。这和它之间是否存在差异？
const Apple& operator =（const Apple&）
除了第二种形式意味着你不能这样做：
Apple a，b，c;
/ /假设有一个功能''Apple运营商+（const Apple&，const
Apple&）''
（a + b）= c; //我认为不允许使用第二个赋值运算符。
Hi,
When a class Apple is written and the assignment operator is not
explicitly declared, the operator
Apple& operator=(const Apple&)
is created by the compiler. Is there any difference between this and
const Apple& operator=(const Apple&)
other than the second form means that you can''t do this:
Apple a, b, c;
//Assume there is a function ''Apple operator+(const Apple&, const
Apple&)''
(a + b) = c; //not allowed with second assignment operator, I think.
任何形式的赋值运算符都不允许这样做，假设
a运营商的有效声明+。
不同的是这个
（a = b）= c
使用第二种形式（a = b）返回一个const引用，然后在另一个运算符的左侧返回
=（因为operator =不是const
方法）。
（a = b）= c对于像int这样的内置类型是没用的，但它是合法的，
所以为什么不呢同样的事情对于用户定义的类也是合法的吗？
john
No this would be allowed by either form of the assignment operator, assuming
a valid declaration of operator+.
The difference is this
(a = b) = c
With the second form (a = b) returns a const reference which cannot then go
on the left hand side of another operator= (because operator= is not a const
method).
(a = b) = c is pretty useless for built in types like int, but it is legal,
so why not make the same thing legal for user defined classes as well?
john
"马修波尔德 <毫安*************** @ kodak.com>在留言中写道
新闻：40 *************** @ kodak.com ...
"Matthew Polder" <ma***************@kodak.com> wrote in message
news:40***************@kodak.com...
当编写Apple类并且未明确声明赋值运算符时，运算符
Apple& operator =（const Apple&）
由编译器创建。这和它之间是否存在差异？
const Apple& operator =（const Apple&）
除了第二种形式意味着你不能这样做：
Apple a，b，c;
/ /假设有一个功能''Apple运营商+（const Apple&，const
Apple&）''
Hi,
When a class Apple is written and the assignment operator is not
explicitly declared, the operator
Apple& operator=(const Apple&)
is created by the compiler. Is there any difference between this and
const Apple& operator=(const Apple&)
other than the second form means that you can''t do this:
Apple a, b, c;
//Assume there is a function ''Apple operator+(const Apple&, const
Apple&)''
从右到左分配运营商组。 />
a = b = c = d，我希望a和d具有相同的值。是否有可能返回
类型是const Apple& ？
Assignment operator groups right to left.
a=b=c=d, i would expect a and d to have same values. Was it possible if return
type was const Apple& ?
" Sharad Kala" <无***************** @ yahoo.com>在消息中写道
新闻：2j ************* @ uni-berlin.de ...
"Sharad Kala" <no*****************@yahoo.com> wrote in message
news:2j*************@uni-berlin.de...
a = b = c = d，我希望a和d具有相同的值。如果返回
类型是const Apple& ？
a=b=c=d, i would expect a and d to have same values. Was it possible if return
type was const Apple& ?
对不起，实际应该是（（a = b）= c）= d。
Sorry, actually it should be ((a=b)=c)=d.
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