Re: Explicit specialization [ Template ]
* AY:
Hi there,
I'm trying to Override a generic version of swapvalue for char. But it
still call's the generic version. Here is the sample code -
template<typename T>
void swapvalues(T& src, T& dest)
{
T temp;
temp = src;
src = dest;
dest = temp;
cout<<"Inside swapvalue [template]"<<endl;
}
// Override of generic version.
void swapvalues(char& src, char& dest)
{
/*
char* temp = NULL;
*temp = src;
src = dest;
dest = *temp;
*/
cout<<"Inside swapvalue [char]"<<endl;
cout<<"src = "<<src<<" : dest = "<<dest<<endl;
}
int main()
{
int i=5, j = 7;
char* srcCh = "ABC";
char* destCh = "XYZ";
Note: you really should make those pointers type 'char const*'. The conversion
from literal to 'char*' is just a C compatibility feature. Up the compiler's
warning level so that it warns about this (e.g. with g++ '-Wwrite-strings').
swapvalues(srcCh, destCh); // Template version is called, despite
address being passed.
// Isn't this the correct way of passing char* to char& ?
//swapvalues(*srcCh, *destCh); // Rt way of calling
cout<<srcCh<<" : "<<destCh<<endl;
return 0;
}
I thought this was the correct way of assigning char* to char&.
char* srcCh = "ABC"
char& refCh = *srcCh; // OK ! Reference is pointing to "ABC";
A tiny hint [ or if time permits a little elaboration ] on why the
explicit specialization is ignored is greatly appreciated.
Technical nitpick: it's not a template specialization, it's an ordinary
overload. But that does not matter here. :-)
It's not ignored.
The first call to swapvalues calls it with pointer arguments, type 'char*'. Only
the template can handle that argument type. The second call, which you have
commented out, calls it with 'char&' type arguments, and you have an overload
for that, which will be called because it competes with the template in being an
exact match, and (rule of thumb, not standard's rule) it's more specific so it wins.
Cheers & hth.,
- Alf
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