Simplify large number of template specializations - c++

Question

So I have a tremendous number of template specializations of this template:
template <typename T> // Same
struct foo { // Same
using type_name = T; // Same
foo(const int base) : _base(base) {} // May take other parameters
void func(const T& param) {} // This function signature will be the same but body will differ
int _base; // Same but may have more members
}; // Same
So an example specialization would be:
template<>
struct foo<float> {
using type_name = T;
foo(const int base, const int child) : _base(base), _child(child) {}
void func(const T& param) { cout << param * _child << endl; }
int _base;
int _child;
};
Obviously this is a toy example and the body of _func will be more involved. But I think this expresses the idea. I can obviously make a macro to help with the boilerplate and put the implementation of the specialized version of the function in an implementation file.
But I was hoping that C++ provided me a way to do this without macros. Is there another way for me avoid writing the boilerplate over and over?

Answer 1

you can have multiple specialization for the function but not for the whole class
like this
#include <iostream>
#include <string>
template<typename T>
struct foo {
//common generic code
using type_name = T;
foo(const int base, const int child) : _base(base), _child(child) {}
void func(const T& param);
int _base;
int _child;
};
template<>
void foo<float>::func(const type_name&) {
//implementation
std::cout << "float" << std::endl;
}
template<>
void foo<int>::func(const type_name&) {
//implementation
std::cout << "int" << std::endl;
}
int main() {
foo<int> tint(0, 0);
foo<float> fint(0, 0);
tint.func(0);
fint.func(0);
}

Answer 2

You can use some light inheritance of data structs to help you separate the differences in member layout and constructor definitions from the main template.
//Define an internal aggregate type you can specialize for your various template parameters
template <typename T>
struct foo_data {
foo(const int base) : _base(base) {}
int _base;
};
//Then derive privately from the data struct (or publicly if you really desire)
template <typename T>
struct foo : private foo_data<T> {
using type_name = T;
using foo_data<T>::foo_data<T>; //Make the base class constructors visible
void func(const T& param); //Use member specialization as suggested by the other answer
};
I will leave it to you to decide if it is better this way or not, but the upshot is that all the common parts are completely separated from all the uncommon parts.
In a comment under another answer I erroneously described this as CRTP. It isn't and it doesn't have any of the drawbacks as CRTP.
If you really need to preserve standard layout, then you can simulate inheritance manually with explicit delegation and perfect forwarding.
template <typename T>
struct foo {
using type_name = T;
template <typename... Args>
foo(Args&&... args) : base_data_(std::forward<Args>(args)...) {}
void func(const T& param); //Use member specialization as suggested by the other answer
foo_data<T> base_data_;
};
One drawback is I don't think the delegating constructor will SFINAE properly as written, and it also eats noexcept specifiers and explicit. Fixing those issues(if required) is left as an exercise to the reader.

Answer 3

There is no nice way to avoid some redundancy in notation when implementing specializations of templated types. There are some techniques to avoid duplication of actual code, such as
Using a traits template to provide type-specific things
template<typename T>
struct foo_traits { ... }; // provide many specialisations
template<typename T> // no specialisations
struct foo
{
using traits = foo_traits<T>;
template<typename...Aars>
explicit foo(Args&&...args)
: data(std::forward<Args>(args)...) {}
int do_something_specific(T x)
{ return traits::do_something(data,x); }
private:
typename traits::data data;
};
a very similar approach is to use a specialized base class:
template<typename T>
struct foo_base { ... }; // provide many specialisations
template<typename T> // no specialisations
struct foo : foo_base<T>
{
using base = foo_base<T>;
template<typename...Aars>
explicit foo(int m, Args&&...args)
: base(std::forward<Args>(args)...)
, more_data(m) {}
int do_something_specific(T x)
{ return base::do_something(x,more_data); }
private:
int more_data;
};
The constructor of foo is a variadic template in order to allow the base class's constructor to take any number and type of arguments.
Of you can use a common base class and specialize the derived classes. This can be done with the Curiously recurring template pattern (CRTP)
template<typename Derived>
struct foo_base // no specializations
{
using type = typename Derived::type;
int do_something(type x)
{
auto result = static_cast<Derived*>(this)->specific_method(x);
return do_some_common_stuff(result);
}
protected:
foo_base(type x) : data(x) {}
type data;
private:
int do_some_common_stuff(type x)
{ /* ... */ }
};
template<typename T> // some specialisations
struct foo : foo_base<foo<T>>
{
using base = foo_base<foo>;
using type = T;
using common_type = typename base::common_type;
using base::do_something;
explicit foo(type x, type y)
: base(x), extra_data(y) {}
protected:
type specific_method(type x)
{ /* ... */ }
private:
type extra_data;
};
Note that foo_base is already a template (unlike the situation with ordinary polymorphism), so you can do a lot of specific stuff there already. Only things that are done differently (not merely with different types) need specializations of foo.
Finally, you can combine these approaches, for example traits classes with CRTP.
All these methods implement some type of static or compile-time polymorphism, rather than real or dynamic polymorphism: there are no virtual functions and hence no virtual table and no overhead for table look-up. It is all resolved at compile time.

Answer 4

This is usually done through inheritance - you put the immutable part into base class, and specialize the children.
I do not think you need an example for that, but let me know if you do.