Hello all,
The code shown below is adapted from Mr. Sutter's Guru of the Week #49.
The 8'th definition of f() is a specialization. What is it a specialization
of (i.e., which is the primary template)???
Is there an answer? If so, is it moot or is the answer meaningful in some
functional way?
Thanks,
Dave
template<class T1, class T2> int f(T1, T2) {return 1;}
template<class T> int f(T) {return 2;}
template<class T> int f(T, T) {return 3;}
template<class T> int f(T*) {return 4;}
template<class T> int f(T*, T) {return 5;}
template<class T> int f(T, T*) {return 6;}
template<class T> int f(int, T*) {return 7;}
template<> int f<int>(int) {return 8;}
int f(int, double) {return 9;}
int f(int) {return 10;} 1 1298
"Dave Theese" <ch**********@yahoo.com> wrote... The code shown below is adapted from Mr. Sutter's Guru of the Week #49.
The 8'th definition of f() is a specialization. What is it a
specialization of (i.e., which is the primary template)???
Is there an answer? If so, is it moot or is the answer meaningful in some functional way?
Thanks, Dave
template<class T1, class T2> int f(T1, T2) {return 1;}
template<class T> int f(T) {return 2;}
template<class T> int f(T, T) {return 3;}
template<class T> int f(T*) {return 4;}
template<class T> int f(T*, T) {return 5;}
template<class T> int f(T, T*) {return 6;}
template<class T> int f(int, T*) {return 7;}
template<> int f<int>(int) {return 8;}
int f(int, double) {return 9;}
int f(int) {return 10;}
I believe (8) is a specialisation of (2). That's the only template
which has one type argument and whose sole function argument is a pure
type as well (not a pointer) and the same as the template type argument.
Wasn't there an answer given to that GotW? Search Google news archives,
I bet it has been discussed in comp.lang.c++.moderated...
Victor This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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