C++ template specialization for pointer? - c++

Question

I read book < C++ Templates - the Complete Guide > and learned template specialization for pointer. (maybe I misunderstand this part of the book)
(1) Here's my simple template:
#include <iostream>
template<typename T>
void Function(const T& a)
{
std::cout << "Function<T>: " << a << std::endl;
}
template<typename T>
void Function<T*>(const T* a)
{
std::cout << "Function<T*>: " << a << std::endl;
}
int main(void)
{
Function(1);
Function(1.2);
Function("hello");
Function((void*)0x25);
return 0;
}
I use ubuntu16.04 x64, g++ 5.3, the compiler report:
$ g++ main.cpp -o main.exe
main.cpp:10:29: error: non-type partial specialization ‘Function<T*>’ is not allowed
void Function<T*>(const T* a)
(2) but this code is correct:
#include <iostream>
template<typename T>
void Function(const T& a)
{
std::cout << "Function<T>: " << a << std::endl;
}
int main(void)
{
Function(1);
Function(1.2);
Function("hello");
Function((void*)0x25);
return 0;
}
result shows:
$ g++ main.cpp -o main.exe
$ ./main.exe
Function<T>: 1
Function<T>: 1.2
Function<T>: hello
Function<T>: 0x25
My question is: Is the book about pointer specialization is wrong ? Or I mis understand the meaning of this part in the book ? Or something else ?
Update about pointer specialization in class.
(3) template class with pointer specialization:
#include <iostream>
template<typename T>
struct Base {
T member;
Base(const T& a)
: member(a)
{
}
void hello()
{
std::cout << member << std::endl;
}
};
template<typename T>
struct Base<T*> {
T* member;
Base(T* a)
: member(a)
{
}
void hello()
{
std::cout << member << std::endl;
}
};
int main(void)
{
Base<int> b1(12);
Base<double> b2(2.4);
Base<char*> b3("hello");
Base<void*> b4((void*)0x25);
b1.hello();
b2.hello();
b3.hello();
b4.hello();
return 0;
}
this code is correct with one warning:
$ g++ main.cpp -o main.exe
main.cpp: In function ‘int main()’:
main.cpp:37:27: warning: deprecated conversion from string constant to ‘char*’ [-Wwrite-strings]
Base<char*> b3("hello");
^
$ ./main.exe
12
2.4
hello
0x25
(4) template class without pointer specialization:
#include <iostream>
template<typename T>
struct Base {
T member;
Base(const T& a)
: member(a)
{
}
void hello()
{
std::cout << member << std::endl;
}
};
int main(void)
{
Base<int> b1(12);
Base<double> b2(2.4);
Base<char*> b3("hello");
Base<void*> b4((void*)0x25);
b1.hello();
b2.hello();
b3.hello();
b4.hello();
return 0;
}
result is the same:
$ g++ main.cpp -o main.exe
main.cpp: In function ‘int main()’:
main.cpp:39:27: warning: deprecated conversion from string constant to ‘char*’ [-Wwrite-strings]
Base<char*> b3("hello");
^
$ ./main.exe
12
2.4
hello
0x25
Does this means pointer specialization is needless ?
Or maybe this feature behave differently on different compiler ?

Answer 1

as you've been already told, partial specialization of function templates are not allowed. You can use std::enable_if for this:
template <typename T, typename std::enable_if_t<!std::is_pointer<T>::value>* = 0>
void func(T val) { std::cout << val << std::endl; }
template <typename T, typename std::enable_if_t<std::is_pointer<T>::value>* = 0>
void func(T val) { func(*val); }
If you are looking for simpler syntax, wait for concepts

Answer 2

The error message told you what is wrong:
non-type partial specialization ‘Function<T*>’ is not allowed
You can only partially specialize types (classes). You've tried to partially specialize a function. Functions are not types; you can only fully specialize them.

Answer 3

Two problems:
You are not allowed to partially specialise a function.
The behaviour of (void*)0x25 is undefined. With the exception of nullptr, you are not allowed to set a pointer to memory you don't own, with the exception of one past the final element of an array and one past the address of a scalar.