I have the below code to understand function full specialization concept:
//Function Full specialization is legal but not partial
class Wrapper
{
public:
void setValue(int x) { }
};
template <typename R, typename T>
R* create(T t)
{
return new R(t);
}
template <>
Wrapper* create<Wrapper, int>(int n) // fully specialized now -> legal...
{
Wrapper* w = new Wrapper();
w->setValue(n);
return w;
}
//template <typename T>
//Wrapper* create<T, int>(T n) // partial specialized now -> illegal...
//{
// Wrapper* w = new Wrapper();
// w->setValue(n);
// return w;
//}
//T
int main()
{
create< Wrapper, int>(2);
create< int, int>(2);
}
The above code compiles and execute fine as expected but when I change the full specialization function signature to something else:
template <>
const char* create<const char*, int>(int n) // fully specialized now -> legal...
{
//Wrapper* w = new Wrapper();
//w->setValue(n);
//return w;
return "Hi";
}
OR
template <>
char* create<char, char>(int n) // fully specialized now -> legal...
{
return (char*)"HI";
}
Error:
explicit specialization 'const char *create<const char*,int>(int)' is not a specialization of a function template Specialization and Overloading
explicit specialization 'char *create<char,char>(int)' is not a specialization of a function template Specialization and Overloading
Why is the error being reported by code and how to fix the same?
template <>
char* create<char, char>(int n)
{
return (char*)"HI";
}
And
template <>
const char* create<const char*, int>(int n)
{
return "Hi";
}
Are not template specializations: the former doesn't have a conforming argument and the later a conforming return type.
template<>
char* create<char, char>(char n) // char n instead of int n
{
return (char*)"HI";
}
Here's a possible template specialization:
template <>
const char** create<const char*, int>(int n) // const char** instead of const char*
{
static const char* test="Hi";
return &test;
}