为什么从std :: map读取不被视为const？ [英] Why does reading from a std::map not considered const?
问题描述
为什么读取std :: map的成员不被视为const？例如：
class My_Class
{
int Get_Map_Value（int index）const // **错误**未考虑
const !!!
{
返回m_Map [index]; //注意m_Map没有改变，只读
来自
}
std :: map< int，int> m_Map;
};
如果尝试将const用于其预期目的，这真的是一团糟......
[== Peteroid ==]
PS - 我正在使用MS VC ++ .NET并创建一个托管C ++应用程序，如果那样
事情......
Peteroid写道：
为什么要阅读std的成员::地图不算const？例如：
类My_Class
{get /Map_Value（int index）const // **错误**不考虑
const !!!
{
返回m_Map [index]; //注意m_Map没有改变，只读
}
std :: map< int，int> m_Map;
};
这真的是在试图将const用于其预期目的时肆虐......
[== Peteroid ==]
PS - 我正在使用MS VC ++ .NET并创建一个托管C ++应用程序，如果这很重要......
如果地图中没有该关键字，map :: operator []将创建一个
新（K，V）对并将其存储在地图中，通过引用返回V部分。
（使用此运算符要求mapped_type是默认构造的。）
您可以通过使用map :: find来避免此行为，尽管您的函数如上面指定的
要么必须返回默认值，要么抛出
异常，如果该函数返回end（）。
-
Doug Harrison
Microsoft MVP - Visual C ++
所以，如果我说得对，那就用std :: map上的operator []不是const，因为
它可能给出的可能性钥匙还没有出现在地图上。
因此，即使我的代码确保我也不会要求输入一个钥匙
还不存在，它不能被声明为const因为我可能不会在这方面写好
好的代码。
我相信我现在明白了，但觉得它应该一直是这样的情况，
operator []被设计为允许const并且如果
请求没有密钥存在的条目，则只会导致错误，并且迫使人们使用count（）方法检查是否存在（如果条目
不存在则返回0，如果条目确实存在，则返回1）。就是这样对待它就像试图
访问一个超出界限的数组元素...
谢谢道格！
[== Peteroid ==]
" Doug Harrison [MVP]" < ds*@mvps.org>在消息中写道
news：cr ******************************** @ 4ax.com ...Peteroid写道：
为什么读取std :: map的成员不被视为const？对于
示例：
类My_Class
{get /Map_Value（int index）const // **错误**不考虑
const !!!
{
返回m_Map [index]; //注意m_Map没有改变，只读
}
std :: map< int，int> m_Map;
};
这真的是在试图将const用于其预期的
目的时肆虐？
[== Peteroid ==]
PS - 我正在使用MS VC ++ .NET并创建一个托管C ++应用程序，如果
那么...
如果密钥是'n'地图中已经存在，map :: operator []将创建一个新的（K，V）对并将其存储在地图中，通过引用返回V部分。
（使用此运算符要求mapped_type是默认构造的。）
你可以通过使用map :: find来避免这种行为，尽管你上面指定的函数要么必须返回默认值，要么抛出如果该函数返回end（），则会出现异常。
-
Doug Harrison
Microsoft MVP - Visual C ++
Peteroid写道：
所以，如果我说得对，在std :: map上使用operator []不是const，因为
它可能给予的可能性给予一张尚未出现在地图中的钥匙。
因此，即使我的代码确保我也不会要求输入一个钥匙
尚未存在的条目，它可以'因为我可能不会在这方面写出好的代码而被声明为const。
我相信我现在明白了，但感觉应该是
operator []的情况被设计为允许const并且如果对没有密钥存在的条目的请求请求只是导致错误，并强制人们使用count（）方法检查是否存在（返回0条目
不存在，如果条目存在，则为1。就是这样，就像试图访问一个超出范围的数组元素一样......
const是C ++中的堕胎。 />
从理论上讲，这听起来是个好主意。在实践中，它只是不起作用。
Why does reading a member of a std::map not considered const? For example:
class My_Class
{
int Get_Map_Value( int index ) const // ** error ** not considered
const!!!
{
return m_Map[index] ; // note that m_Map is not changed, only read
from
}
std::map<int,int> m_Map ;
} ;
This really reeks havoc when to try to use const for its intended purpose...
[==Peteroid==]
PS - I''m using MS VC++.NET and creating a Managed C++ application, if that
matters...
Peteroid wrote:
Why does reading a member of a std::map not considered const? For example:
class My_Class
{
int Get_Map_Value( int index ) const // ** error ** not considered
const!!!
{
return m_Map[index] ; // note that m_Map is not changed, only read
from
}
std::map<int,int> m_Map ;
} ;
This really reeks havoc when to try to use const for its intended purpose...
[==Peteroid==]
PS - I''m using MS VC++.NET and creating a Managed C++ application, if that
matters...
If the key isn''t already present in the map, map::operator[] will create a
new (K,V) pair and store it in the map, returning the V part by reference.
(Use of this operator requires the mapped_type be default-constructible.)
You can avoid this behavior by using map::find, though your function as
specified above will either have to return a default value or throw an
exception if that function returns end().
--
Doug Harrison
Microsoft MVP - Visual C++
So, if I get you right, using operator[] on a std::map isn''t const because
of the POSSIBILITY it might give be given a key not already in the map.
Thus, even if my code insures I won''t ever ask for an entry for which a key
doesn''t already exist, it can''t be declared const since I might not write
good code in this regard.
I believe I now understand, but feel it should have been the case that
operator[] was designed to allow const and to just cause an error if a
request for an entry where no key existed was asked for, and force people to
check on existence using the count() method (which returns 0 if the entry
doesn''t exist and 1 if it does). That is, treat it much like trying to
access an array element that is out of bounds...
Thanks Doug!
[==Peteroid==]
"Doug Harrison [MVP]" <ds*@mvps.org> wrote in message
news:cr********************************@4ax.com...Peteroid wrote:Why does reading a member of a std::map not considered const? For example:
class My_Class
{
int Get_Map_Value( int index ) const // ** error ** not considered
const!!!
{
return m_Map[index] ; // note that m_Map is not changed, only read
from
}
std::map<int,int> m_Map ;
} ;
This really reeks havoc when to try to use const for its intended purpose...
[==Peteroid==]
PS - I''m using MS VC++.NET and creating a Managed C++ application, if thatmatters...
If the key isn''t already present in the map, map::operator[] will create a
new (K,V) pair and store it in the map, returning the V part by reference.
(Use of this operator requires the mapped_type be default-constructible.)
You can avoid this behavior by using map::find, though your function as
specified above will either have to return a default value or throw an
exception if that function returns end().
--
Doug Harrison
Microsoft MVP - Visual C++
Peteroid wrote:
So, if I get you right, using operator[] on a std::map isn''t const because
of the POSSIBILITY it might give be given a key not already in the map.
Thus, even if my code insures I won''t ever ask for an entry for which a key
doesn''t already exist, it can''t be declared const since I might not write
good code in this regard.
I believe I now understand, but feel it should have been the case that
operator[] was designed to allow const and to just cause an error if a
request for an entry where no key existed was asked for, and force people to
check on existence using the count() method (which returns 0 if the entry
doesn''t exist and 1 if it does). That is, treat it much like trying to
access an array element that is out of bounds...
const is an abortion in C++.
In theory, it sounds like a good idea. In practice, it just doesn''t work.
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