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Hi:
When compiling the folowing code under visual c++ 2005 I get he following
error:
C3848: expression having type 'const BitSetComp' would lose some
const-volatile qualifiers in order to call 'const bool BitSetComp:perator
()(const KeyType &,const KeyType &)'
This is in the file xtree on line 1174. The code causing this is:
struct BitSetComp

{

bool const operator()(const KeyType& lhs, const KeyType& rhs)

{

return lhs.to_ulong() < rhs.to_ulong();

}

};

Anyone got a clue what the issue is, and what's causing this?

What is the solution?

Cheers

Sean.

Paavo Helde wrote:
> "Sean Farrow" <> kirjutas:
>
>> Hi:
>> When compiling the folowing code under visual c++ 2005 I get he
>> following error:
>> C3848: expression having type 'const BitSetComp' would lose some
>> const-volatile qualifiers in order to call 'const bool
>> BitSetComp:perator ()(const KeyType &,const KeyType &)'
>> This is in the file xtree on line 1174. The code causing this is:
>> struct BitSetComp
>>
>> {
>>
>> bool const operator()(const KeyType& lhs, const KeyType& rhs)
>
> Not enough info, but try:
>
> bool operator() const (const KeyType& lhs, const KeyType& rhs)
>
Or if you want it to compile,

bool operator()(const KeyType& lhs, const KeyType& rhs) const

--
Ian Collins.

Hi:
Thanks both, compiles successfully.
Sean.
"Ian Collins" <ian-> wrote in message
news:...
> Paavo Helde wrote:
>> "Sean Farrow" <> kirjutas:
>>
>>> Hi:
>>> When compiling the folowing code under visual c++ 2005 I get he
>>> following error:
>>> C3848: expression having type 'const BitSetComp' would lose some
>>> const-volatile qualifiers in order to call 'const bool
>>> BitSetComp:perator ()(const KeyType &,const KeyType &)'
>>> This is in the file xtree on line 1174. The code causing this is:
>>> struct BitSetComp
>>>
>>> {
>>>
>>> bool const operator()(const KeyType& lhs, const KeyType& rhs)
>>
>> Not enough info, but try:
>>
>> bool operator() const (const KeyType& lhs, const KeyType& rhs)
>>
> Or if you want it to compile,
>
> bool operator()(const KeyType& lhs, const KeyType& rhs) const
>
> --
> Ian Collins.

"Ian Collins"
> Or if you want it to compile,
>
> bool operator()(const KeyType& lhs, const KeyType& rhs) const
>


Is there any reason to make one non-constant? A functor wouldn't be
adaptable if it wasn't constant.


Fraser.