Hello,
I wrote an (templatized) operator==() and don't know why the compiler
doesn't consider it. What prevents it from being chosen?
class Data {};
template <typename D>
class Vector
{
public:
typedef Data ConstType;
typedef Data Type;
};
template <typename D>
bool
operator==(cons t typename Vector<D>::Cons tType &lhs,
const typename Vector<D>::Type &rhs)
{
return true;
}
// this operator is chosen, if added.
//bool
//operator==(cons t Vector<Data>::C onstType &lhs,
// const Vector<Data>::T ype &rhs)
//{
// return true;
//}
int
main()
{
Vector<Data>::C onstType ct;
Vector<Data>::T ype t;
ct == t;
return 0;
}
What is not standard conforming, what is not allowed here?
regards,
alex 5 1366
"Alexander Stippler" <st**@mathemati k.uni-ulm.de> wrote in message
news:3f******@n ews.uni-ulm.de... Hello,
I wrote an (templatized) operator==() and don't know why the compiler doesn't consider it. What prevents it from being chosen?
class Data {};
template <typename D> class Vector { public: typedef Data ConstType; typedef Data Type; };
template <typename D> bool operator==(cons t typename Vector<D>::Cons tType &lhs, const typename Vector<D>::Type &rhs) { return true; }
// this operator is chosen, if added. //bool //operator==(cons t Vector<Data>::C onstType &lhs, // const Vector<Data>::T ype &rhs) //{ // return true; //}
int main() { Vector<Data>::C onstType ct; Vector<Data>::T ype t;
ct == t;
return 0; }
What is not standard conforming, what is not allowed here?
regards, alex
Do you mean to say that your template operator==() is not chosen even if the
non-template version is commented out? Or do you mean to say that if the
non-template version is not commented out that it is preferred over the
template version? If the latter is the case, that makes sense since
non-templates are always preferred over templates...
Dave wrote: Do you mean to say that your template operator==() is not chosen even if the non-template version is commented out? Or do you mean to say that if the non-template version is not commented out that it is preferred over the template version? If the latter is the case, that makes sense since non-templates are always preferred over templates...
Yes. That's absolutely clear. But I mean the first. If I have only the
template version, it is not chosen!
"Alexander Stippler" <st**@mathemati k.uni-ulm.de> wrote in message
news:3f******@n ews.uni-ulm.de... Hello,
I wrote an (templatized) operator==() and don't know why the compiler doesn't consider it. What prevents it from being chosen?
class Data {};
template <typename D> class Vector { public: typedef Data ConstType; typedef Data Type;
I assume that you've missed out typedef D Data?
why are Type and ConstType the same?
surely typedef const Data ConstType.
};
template <typename D> bool operator==(cons t typename Vector<D>::Cons tType &lhs, const typename Vector<D>::Type &rhs) { return true; }
I think that someone else recently posted a v similar question.
I don't know why the compiler doesn't consider it but I do know that even
attempting this is logically the wrong way to go for a number of reasons:
1. typedefs do not intoduce new types - they are only aliases for existing
types therefore all instantiations actually have the same signature i.e.
operator==(cons t Data&,const Data&) i.e. you cannot distinguish between
equality of Data when used in a Vector and when used in any other way.
2. The equality of Data is logically a part of the interface of Data not of
Vector - it is not the responsibility of a vector to define operations for
its parameters.
3. The whole idea is fundamantally wrong - If Data is a parameter and hence
you don't know what it is then you cannot logically define equality for it.
// this operator is chosen, if added. //bool //operator==(cons t Vector<Data>::C onstType &lhs, // const Vector<Data>::T ype &rhs) //{ // return true; //}
int main() { Vector<Data>::C onstType ct; Vector<Data>::T ype t;
ct == t;
return 0; }
What is not standard conforming, what is not allowed here?
regards, alex
"Alexander Stippler" <st**@mathemati k.uni-ulm.de> wrote in message
news:3f******@n ews.uni-ulm.de... Hello,
I wrote an (templatized) operator==() and don't know why the compiler doesn't consider it. What prevents it from being chosen?
class Data {};
template <typename D> class Vector { public: typedef Data ConstType; typedef Data Type; };
template <typename D> bool operator==(cons t typename Vector<D>::Cons tType &lhs, const typename Vector<D>::Type &rhs) { return true; }
// this operator is chosen, if added. //bool //operator==(cons t Vector<Data>::C onstType &lhs, // const Vector<Data>::T ype &rhs) //{ // return true; //}
int main() { Vector<Data>::C onstType ct; Vector<Data>::T ype t;
ct == t;
return 0; }
What is not standard conforming, what is not allowed here?
regards, alex
Here's the best I could find:
On page 170 of "C++ Templates", the following bullet item occurs and lists
one particular construct that is not a so-called "deduced context":
"Qualified type names. A type name like Q<T>::X will never be used to
deduce a template parameter T, for example."
You're absolutely right. That obvious yet that hidden.
I will rethink the design. That probably even makes things easier :-).
Thanks! This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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