奇怪的错误 [英] weird error

问题描述

（更改文件和类名以保护无辜者）


g ++ -Wall -c file.cpp：在成员函数`std :: string

Some_Class :: get_member（const std :: string&）const''：

file.cpp：46：传递`const std :: map< std :: string，

std :: string，std :: less< std :: string>，std :: allocator< std :: pair< const

std :: string，std :: string> > >''作为`_Tp&的'this'参数std :: map< _Key，

_Tp，_Compare，_ Alloc> :: operator []（const _Key&）[with _Key = std :: string，

_Tp = std :: string，_Compare = std :: less< std :: string>，_ Alloc =

std :: allocator< std :: pair< const std :: string，std :: string> >]''丢弃

限定符


我无法做出这个错误的正面或反面。我猜它在某些const对象上调用一个非const函数时会有一些东西。任何想法？


函数SomeClass :: get_member是：


std :: string SomeClass :: get_member（const std :: string& ;姓名）const

{

返回_member [姓名];

}


其中_成员是一个std :: map< std :: string，std :: string> ;.

(file and class names changed to protect the innocent)

g++ -Wall -c file.cpp: In member function `std::string
Some_Class::get_member(const std::string&) const'':
file.cpp:46: passing `const std::map<std::string,
std::string, std::less<std::string>, std::allocator<std::pair<const
std::string, std::string> > >'' as `this'' argument of `_Tp& std::map<_Key,
_Tp, _Compare, _Alloc>::operator[](const _Key&) [with _Key = std::string,
_Tp = std::string, _Compare = std::less<std::string>, _Alloc =
std::allocator<std::pair<const std::string, std::string> >]'' discards
qualifiers

I can''t make heads or tails of this error. I''m guessing it has something to
do with calling a non-const function on some const object. Any ideas?

The function SomeClass::get_member is:

std::string SomeClass::get_member(const std::string &name) const
{
return _member[name];
}

where _member is a std::map<std::string, std::string>.

推荐答案

" Joe Laughlin" <乔*************** @ boeing.com>写道...

"Joe Laughlin" <Jo***************@boeing.com> wrote...

（文件和类名改变以保护无辜者）

g ++ -Wall -c file.cpp：在成员函数`std :: string
Some_Class :: get_member（const std :: string&）const''：
file.cpp：46：传递`const std :: map< std :: string，
std :: string， std :: less< std :: string>，std :: allocator< std :: pair< const
std :: string，std :: string> > >''作为`_Tp&
std :: map< _Key，
_Tp，_Compare，_Alloc> :: operator []（const _Key&）的'this'参数[与_Key =
std :: string，
_Tp = std :: string，_Compare = std :: less< std :: string>，_ Alloc =
std :: allocator< std :: pair< const std :: string，std :: string> >]''丢弃
限定符

我无法做出这个错误的正面或反面。我猜它在某些const对象上调用非const函数时会有什么作用。任何想法？

函数SomeClass :: get_member是：

std :: string SomeClass :: get_member（const std :: string& name）const
{
返回_member [name];
}
其中_member是std :: map< std :: string，std :: string>。

(file and class names changed to protect the innocent)

g++ -Wall -c file.cpp: In member function `std::string
Some_Class::get_member(const std::string&) const'':
file.cpp:46: passing `const std::map<std::string,
std::string, std::less<std::string>, std::allocator<std::pair<const
std::string, std::string> > >'' as `this'' argument of `_Tp&
std::map<_Key,
_Tp, _Compare, _Alloc>::operator[](const _Key&) [with _Key =
std::string,
_Tp = std::string, _Compare = std::less<std::string>, _Alloc =
std::allocator<std::pair<const std::string, std::string> >]'' discards
qualifiers

I can''t make heads or tails of this error. I''m guessing it has something
to
do with calling a non-const function on some const object. Any ideas?

The function SomeClass::get_member is:

std::string SomeClass::get_member(const std::string &name) const
{
return _member[name];
}

where _member is a std::map<std::string, std::string>.

您声明了此成员函数（''get_member''）" const"。这意味着

对象（除了''mutable''之外的所有数据成员）都不会改变
。 ''map''模板的索引操作符是_non-const_函数。

I.e.它不能被称为const映射。它必须是非const的，因为它可能会插入一个新的对象，如果它还没有那么。


你需要和平事实上它可能会在你的_member中插入另一个

对象（并声明''_member'''可变的'）或使用其他

方式。看来你的函数_assumes_在地图中有一个元素



键''name''。如果没有怎么办？


Victor

You declared this member function (''get_member'') "const". That means that
the object (and all its data member except ''mutable'' ones) are not going to
change. Indexing operator for ''map'' template is a _non-const_ function.
I.e. it cannot be called for a const map. It has to be non-const because
it may insert a new object if it''s not there yet.

You need to either make peace with the fact that it may insert another
object into your _member (and declare ''_member'' "mutable") or use some other
way. It seems that your function _assumes_ that there is an element with
the
key ''name'' in the map. What if there isn''t?

Victor

Joe Laughlin写道：

Joe Laughlin wrote:

（file和类名改变以保护无辜者）

g ++ -Wall -c file.cpp：在成员函数`std :: string
Some_Class :: get_member（const std :: string&） const''：
file.cpp：46：传递`const std :: map< std :: string，
std :: string，std :: less< std :: string>，std :: allocator< std :: pair< const
std :: string，std :: string> > >''作为`_Tp&的'this'参数std :: map< _Key，
_Tp，_Compare，_ Alloc> :: operator []（const _Key&）[with _Key = std :: string，
_Tp = std :: string，_Compare = std :: less< std :: string>，_ Alloc =
std :: allocator< std :: pair< const std :: string，std :: string> >]''丢弃
限定符

我无法做出这个错误的正面或反面。我猜它在某些const对象上调用非const函数时会有所作为。任何想法？

函数SomeClass :: get_member是：

std :: string SomeClass :: get_member（const std :: string& name）const
{
返回_member [姓名];


你的函数被声明为const，但你无法保证地图的

下标运算符不会改变''name''的值。用

换句话说：在const成员函数中，你只能使用其他const

函数（关于你自己的成员）。在上面，如果''name''确实
$ _ $ b在_member中不存在，*它将被创建*。你不能在const函数中做

。

}

(file and class names changed to protect the innocent)

g++ -Wall -c file.cpp: In member function `std::string
Some_Class::get_member(const std::string&) const'':
file.cpp:46: passing `const std::map<std::string,
std::string, std::less<std::string>, std::allocator<std::pair<const
std::string, std::string> > >'' as `this'' argument of `_Tp& std::map<_Key,
_Tp, _Compare, _Alloc>::operator[](const _Key&) [with _Key = std::string,
_Tp = std::string, _Compare = std::less<std::string>, _Alloc =
std::allocator<std::pair<const std::string, std::string> >]'' discards
qualifiers

I can''t make heads or tails of this error. I''m guessing it has something to
do with calling a non-const function on some const object. Any ideas?

The function SomeClass::get_member is:

std::string SomeClass::get_member(const std::string &name) const
{
return _member[name];
Your function is declared const, but you cannot guarantee that the
subscript operator of the map doesn''t change the value at ''name''. In
other words: In a const member function you can only use other const
functions (in regard to your own members). In the above, if ''name'' does
not exist in _member, *it will be created*. You are not allowed to do
that in a const function.
}

试试这个：


std :: string SomeClass :: get_member（std :: string const& name）const

{

_member.const_iterator it ;

it = _member.find（name）;

if（it！= _member.end（））

_return it->第二个;

其他

< ...> //无论你想做什么，如果''名字''不存在......

}


占用更多空间，但它的确有效。在大多数编译器上我应该

认为它也同样有效。


-

/ Brian Riis

try this:

std::string SomeClass::get_member(std::string const & name) const
{
_member.const_iterator it;
it = _member.find(name);
if(it != _member.end())
_return it->second;
else
<...> // Whatever you want to do, if ''name'' isn''t there...
}

Takes up a bit more space, but it works. On most compilers I should
think it is as efficient too.

--
/Brian Riis

Brian Riis写道：

Brian Riis wrote:

Joe Laughlin写道：

Joe Laughlin wrote:

（文件和类名改为保护无辜者）

g ++ -Wall -c file.cpp：在成员函数`std :: string
Some_Class :: get_member（const std :: string&）const''：
file.cpp：46：传递`const std :: map< std :: string，
std :: string，std :: less< std :: string>，
std :: allocator< std :: pair< const std： ：string，
std :: string> > >''作为`_Tp&
std :: map< _Key，_Tp，_Compare，
_Alloc> :: operator []（const _Key&）的'this'参数[与_Key =
std :: string，_比较=
std :: less< std :: string>，_ Alloc =
:: string，std :: string>

(file and class names changed to protect the innocent)

g++ -Wall -c file.cpp: In member function `std::string
Some_Class::get_member(const std::string&) const'':
file.cpp:46: passing `const std::map<std::string,
std::string, std::less<std::string>,
std::allocator<std::pair<const std::string,
std::string> > >'' as `this'' argument of `_Tp&
std::map<_Key, _Tp, _Compare,
_Alloc>::operator[](const _Key&) [with _Key =
std::string, _Tp = std::string, _Compare =
std::less<std::string>, _Alloc =
std::allocator<std::pair<const std::string, std::string>

>]''丢弃限定符

>]'' discards qualifiers

我不能对此错误做出正面或反面。我猜测它与在一些const对象上调用非const函数有关。任何想法？

函数SomeClass :: get_member是：

std :: string SomeClass :: get_member（const std :: string
&name;）const {
返回_member [name];

I can''t make heads or tails of this error. I''m guessing
it has something to do with calling a non-const function
on some const object. Any ideas?

The function SomeClass::get_member is:

std::string SomeClass::get_member(const std::string
&name) const {
return _member[name];

你的函数被声明为const，但你无法保证
地图的下标运算符不会改变
'name''的价值。换句话说：在const成员
函数中，您只能使用其他const函数（关于您自己的成员）。在上面，如果'_ name''在_member中不存在，*它将被创建*。你不能在const函数中这样做。

Your function is declared const, but you cannot guarantee
that the subscript operator of the map doesn''t change the
value at ''name''. In other words: In a const member
function you can only use other const functions (in
regard to your own members). In the above, if ''name'' does
not exist in _member, *it will be created*. You are not
allowed to do that in a const function.

}

试试这个：

std :: string SomeClass :: get_member（std :: string const&
name）const {
_member.const_iterator it;
it = _member.find（name）;
if（it！= _member.end（））
_return it-> second;
else
< ...> //无论你想做什么，如果''名字''不在那里......}

占用更多空间，但它有效。在大多数
编译器上，我认为它也同样有效。

try this:

std::string SomeClass::get_member(std::string const &
name) const {
_member.const_iterator it;
it = _member.find(name);
if(it != _member.end())
_return it->second;
else
<...> // Whatever you want to do, if ''name'' isn''t
there... }

Takes up a bit more space, but it works. On most
compilers I should think it is as efficient too.

谢谢你们。


是：

_member.const_iterator it;

合法？编译器不喜欢那条线，但确实如此：

std :: map< std :: string，std :: string> :: const_iterator it;


Joe

Thanks you guys.

Is:
_member.const_iterator it;
legal? The compiler didn''t like that line, but it did like:
std::map<std::string, std::string>::const_iterator it;

Joe

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