#include < iostream.h >
template < class T>
class JGTuplet
{
friend ostream& operator < <(ostream&, const JGTuplet < T > &);
.
.
.
The rest of the class
};
This would have worked in BCB 4.0 Pro but Linux gcc complains that the above code does not declare a templated friend function. This is the way I've learned how to do it. It suggests having the template function already declared, and placing a < Type > in front of the opening function parenthesses (after the operator<< in the above code). What gives?
Thanks,
joeG
Edited by - joeG on 6/16/00 3:36:03 PM
Templated Friend's???
Author
What would be the proper syntax for declaring a templated friend method? This is what I have so far
I can't say for sure, but in DJGPP you'd write:
to make it work. Maybe worth a shot?
/TP
Edited by - suburber on June 16, 2000 5:45:09 PM
friend ostream& operator<< <> ( ostream&, const JGTuplet<T>& ); to make it work. Maybe worth a shot?
/TP
Edited by - suburber on June 16, 2000 5:45:09 PM
quote:Original post by suburber
I can''t say for sure, but in DJGPP you''d write:friend ostream& operator<< <> ( ostream&, const JGTuplet<T>& ); [/source]to make it work. Maybe worth a shot?/TPEdited by - suburber on June 16, 2000 5:45:09 PM <hr height=1 noshade></BLOCKQUOTE></font> Um, didn''t work. Of what I gathered g++ requires you to do something like this[source] template <class T>friend ostream& operator<< <T> (ostream&, const JGTuplet<T>&);
I''ll try it without the T in the middle of the template specifiers between "operator<<" and "(ostream &..."
joeG
Author
Um, didn''t work. Of what I gathered g++ requires you to do something like this
I''ll try it without the T in the middle of the template specifiers between "operator<<" and "(ostream &..."
joeG
Sorry about how ugly the above post turned out. I wouldn''t mind at all if it got deleted *Hint, hint*
template <class T>friend ostream& operator<< <T> (ostream&, const JGTuplet<T>&); I''ll try it without the T in the middle of the template specifiers between "operator<<" and "(ostream &..."
joeG
Sorry about how ugly the above post turned out. I wouldn''t mind at all if it got deleted *Hint, hint*
Author
Um, finally hacked my way to the solution...
In gcc (g++, c++) you'll have to do something like this.
I hope this will prove helpful to some of the people trying to work with Linux/Unix gcc.
joeG
Edited by - joeG on June 19, 2000 10:00:38 AM
In gcc (g++, c++) you'll have to do something like this.
In file, x.h
template <class Type> class X;template <class Type> ostream& operator<<(ostream&, X<Type>&);template <class Type>class X{private:protected:public: friend ostream& operator<< <Type> (ostream &, const X<Type>&);}; [/source]<hr><h1>In file, x.cpp</h1> [source]template class X<int>;template class X<long>;template class X<float>;template class X<double>;template class X<char*>;template class X<bool>;template ostream& operator<<(ostream&, X<int>);template ostream& operator<<(ostream&, X<long>);template ostream& operator<<(ostream&, X<float>);template ostream& operator<<(ostream&, X<double>);template ostream& operator<<(ostream&, X<char*>);template ostream& operator<<(ostream&, X<bool>);template <class Type>ostream& operator<<(ostream&, X<Type>){ //implementation}[/source]<hr><h1>in file, main.cpp</h1>[source]#include "x.h"int main(int argc, char **argv){ X<int> a; /* The following line would cause an error because you didn't explicitly declare to gcc (or the linker) that you expect a template type of `unsigned char' */ X<unsigned char> b; . . . return 0;} I hope this will prove helpful to some of the people trying to work with Linux/Unix gcc.
joeG
Edited by - joeG on June 19, 2000 10:00:38 AM
Author
I''ve given up trying to fix the above post, I think you all will get the idea, but if one of the moderators would be so kind, I''d be very grateful.
joeG
joeG
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