引用类型方法 [英] reference type methods
问题描述
为什么一个方法被定义为返回一个引用，例如
运算符重载[]，运算符[]，
href& operator []（int index）{
返回_array [index];
}
不会导致类型不匹配编译错误？
Ruben
-
http://www.mrbrklyn.com - 有趣的东西
http://www.nylxs.com - 自由软件的领导力发展
如此多的移民群体席卷布鲁克林，如亚特兰蒂斯，在世界的心灵中达到神话般的比例 - RI Safir 1998
http：/ /fairuse.nylxs.com DRM是盗窃者 - 我们是利益相关者 - RI Safir 2002
是的 - 我写自由软件...所以SUE ME
我们面临的巨大问题是我们正在成为我们自己的文化遗产的佃农 - 我们需要有能力参与我在我们自己的社会中。
我是一名工程师。我选择了最适合工作的工具，政治被诅咒。<
你必须是一个愚蠢的工程师，因为自古埃及第一王朝以来，政治和技术一直在附着。我猜你错过了那个。
??数字千年的版权
Why would a method that is defined to return a reference such as with the
operator overload of [], operator[],
href& operator[](int index){
return _array[index];
}
not cause a type mismatch compiler error?
Ruben
--
http://www.mrbrklyn.com - Interesting Stuff
http://www.nylxs.com - Leadership Development in Free Software
So many immigrant groups have swept through our town that Brooklyn, like Atlantis, reaches mythological proportions in the mind of the world - RI Safir 1998
http://fairuse.nylxs.com DRM is THEFT - We are the STAKEHOLDERS - RI Safir 2002
"Yeah - I write Free Software...so SUE ME"
"The tremendous problem we face is that we are becoming sharecroppers to our own cultural heritage -- we need the ability to participate in our own society."
"I''m an engineer. I choose the best tool for the job, politics be damned.<
You must be a stupid engineer then, because politcs and technology have been attached at the hip since the 1st dynasty in Ancient Egypt. I guess you missed that one."
?? Copyright for the Digital Millennium
推荐答案
Ruben写道：
Ruben wrote:
为什么将一个方法定义为返回一个引用，例如
运算符重载[]，运算符[]，
返回_array [index];
}
不会导致类型不匹配编译错误？
Why would a method that is defined to return a reference such as with the
operator overload of [], operator[],
href& operator[](int index){
return _array[index];
}
not cause a type mismatch compiler error?
为什么要这样？假设_array是一个href数组，它只返回一个对href的
引用。
-
Ian Collins。
Why should it? Assuming _array is an array of href, it just returns a
reference to an href.
--
Ian Collins.
2008年7月14日星期一22:11:19 +1200，Ian Collins写道：
On Mon, 14 Jul 2008 22:11:19 +1200, Ian Collins wrote:
Ruben写道：
Ruben wrote:
>为什么一个方法被定义为返回一个引用，如
运算符重载[]，operator []，
href& operator []（int index）{
返回_array [index];
}
}
不会导致类型不匹配编译器错误？
>Why would a method that is defined to return a reference such as with
the operator overload of [], operator[],
href& operator[](int index){
return _array[index];
}
}
not cause a type mismatch compiler error?
为什么要这样？假设_array是一个href数组，它只返回一个对href的
引用。
Why should it? Assuming _array is an array of href, it just returns a
reference to an href.
hmmm
符号数组不会作为参考
它将被定义
私人：
int _array [default_size];
这来自教科书，我找到了令人费解。
Ruben
-
http://www.mrbrklyn.com - 有趣的东西
http://www.nylxs.com - 自由软件的领导力发展
如此多的移民群体席卷我们的城镇布鲁克林，如亚特兰蒂斯，达到神话世界头脑中的比例 - RI Safir 1998
http：// fairuse .nylxs.com DRM是盗窃者 - 我们是利益相关者 - RI Safir 2002
是的 - 我写自由软件...所以SUE ME
"颤抖我们面临的一个重大问题是，我们正在成为我们自己的文化遗产的佃农 - 我们需要有能力参与我们自己的社会。
我是一名工程师。我选择了最适合工作的工具，政治被诅咒。<
你必须是一个愚蠢的工程师，因为自古埃及第一王朝以来，政治和技术一直在附着。我猜你错过了那个。
??版权所有数字千年
hmmm
the symbol array would not be a reference
it would be defined
private:
int _array[default_size];
This comes up from a text book, and I found it puzzling.
Ruben
--
http://www.mrbrklyn.com - Interesting Stuff
http://www.nylxs.com - Leadership Development in Free Software
So many immigrant groups have swept through our town that Brooklyn, like Atlantis, reaches mythological proportions in the mind of the world - RI Safir 1998
http://fairuse.nylxs.com DRM is THEFT - We are the STAKEHOLDERS - RI Safir 2002
"Yeah - I write Free Software...so SUE ME"
"The tremendous problem we face is that we are becoming sharecroppers to our own cultural heritage -- we need the ability to participate in our own society."
"I''m an engineer. I choose the best tool for the job, politics be damned.<
You must be a stupid engineer then, because politcs and technology have been attached at the hip since the 1st dynasty in Ancient Egypt. I guess you missed that one."
?? Copyright for the Digital Millennium
Ruben写道：
Ruben wrote:
2008年7月14日星期一22 ：11：19 +1200，Ian Collins写道：
On Mon, 14 Jul 2008 22:11:19 +1200, Ian Collins wrote:
> Ruben写道：
>Ruben wrote:
> >为什么一个方法被定义为返回一个引用，例如用[]，operator []，
href&的运算符重载。 operator []（int index）{
返回_array [index];
}
}
不会导致类型不匹配编译器错误？
>>Why would a method that is defined to return a reference such as with
the operator overload of [], operator[],
href& operator[](int index){
return _array[index];
}
}
not cause a type mismatch compiler error?
为什么要这样？假设_array是一个href数组，它只返回一个对href的引用。
Why should it? Assuming _array is an array of href, it just returns a
reference to an href.
hmmm
符号数组不会作为参考
它将被定义
私人：
int _array [default_size];
hmmm
the symbol array would not be a reference
it would be defined
private:
int _array[default_size];
是的，但_array [index]在一个int。我只能猜测href是int的
typedef别名。该函数返回对该int的引用。
-
Ian Collins。
Yes, but _array[index] in an int. I can only guess that href is a
typedef alias for int. The function returns a reference to that int.
--
Ian Collins.
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