C++ template specialization of constructor - c++

Question

I have a templated class A<T, int> and two typedefs A<string, 20> and A<string, 30>.
How do I override the constructor for A<string, 20> ? The following does not work:
template <typename T, int M> class A;
typedef A<std::string, 20> one_type;
typedef A<std::string, 30> second_type;
template <typename T, int M>
class A {
public:
A(int m) {test= (m>M);}
bool test;
};
template<>
one_type::one_type() { cerr << "One type" << endl;}
I would like the class A<std::string,20> to do something that the other class doesn't. How can I do this without changing the constructor A:A(int) ?

Answer 1

The only thing you cannot do is use the typedef to define the constructor. Other than that, you ought to specialize the A<string,20> constructor like this:
template<> A<string,20>::A(int){}
If you want A<string,20> to have a different constructor than the generic A, you need to specialize the whole A<string,20> class:
template<> class A<string,20> {
public:
A(const string& takethistwentytimes) { cerr << "One Type" << std::endl; }
};

Answer 2

Assuming your really meant for A::test to be publicly accessible, you could do something like this:
#include <iostream>
template <int M>
struct ABase
{
ABase(int n) : test_( n > M )
{}
bool const test_;
};
template <typename T, int M>
struct A : ABase<M>
{
A(int n) : ABase<M>(n)
{}
};
template <typename T>
A<T, 20>::A(int n)
: ABase<20>(n)
{ std::cerr << "One type" << std::endl; }
Kick the tires:
int main(int argc, char* argv[])
{
A<int, 20> a(19);
std::cout << "a:" << a.test_ << std::endl;
A<int, 30> b(31);
std::cout << "b:" << b.test_ << std::endl;
return 0;
}

Answer 3

Late but a very elegant solution:
C++ 2020 introduced Constraints and Concepts. You can now conditionally enable and disable constructors and destructors!
#include <iostream>
#include <type_traits>
template<class T>
struct constructor_specialized
{
constructor_specialized() requires(std::is_same_v<T, int>)
{
std::cout << "Specialized Constructor\n";
};
constructor_specialized()
{
std::cout << "Generic Constructor\n";
};
};
int main()
{
constructor_specialized<int> int_constructor;
constructor_specialized<float> float_constructor;
};
Run the code here.

Answer 4

This may be a little bit late, but if you have access to c++11 you can use SFINAE to accomplish just what you want:
template <class = typename std::enable_if<
std::is_same<A<T,M>, A<std::string, 20>>::value>::type // Can be called only on A<std::string, 20>
>
A() {
// Default constructor
}
Working example

Answer 5

How about :
template<typename T, int M, bool dummy = (M > 20) >
class A {
public:
A(int m){
// this is true
}
};
template<typename T, int M>
class A<T,M,false> {
public:
A(int m) {
//something else
}
};

Answer 6

You can't with your current approach. one_type is an alias to a particular template specialization, so it gets whatever code the template has.
If you want to add code specific to one_type, you have to declare it as a subclass of A specialization, like this:
class one_type:
public A<std::string, 20>
{
one_type(int m)
: A<str::string, 20>(m)
{
cerr << "One type" << endl;
}
};

Answer 7

The best solution I've been able to come up with for this situation is to use a "constructor helper function":
template <typename T, int M> class A;
typedef A<std::string, 20> one_type;
typedef A<std::string, 30> second_type;
template <typename T, int M>
class A {
private:
void cons_helper(int m) {test= (m>M);}
public:
A(int m) { cons_helper(m); }
bool test;
};
template <>
void one_type::cons_helper(int) { cerr << "One type" << endl;}