Specialize overloaded constructor by type of parameter C++

Question

template <typename T> class foo
{
private:
    T m_var;
public:
    foo(T init_var)
    {
        m_var = init_var;
    }
    foo(int init_var)
    {
        m_var = init_var + 1;
    }
};

int main()
{
    foo1 = foo<int>(3); // foo1.m_var is 4
    foo2 = foo<char>('a'); // foo2.m_var is a
    return 0;
}

Can I specialize template class constructor like this? Class foo works on types generaly, but when its constructor called, it works diffrently as the type of parameter. I want to use template if I can, but I see the answer which is saying 'Cant use template in constructor'.

Answer 1

You can have a different constructor be called when the template argument type is int by adding a specialization.

template <typename T> 
class foo
{
private:
    T m_var;
public:
    foo(T init_var)
    {
        m_var = init_var;
    }
};

template<>
foo<int>::foo(int init_var)
{
    m_var = init_var + 1;
}

Live demo

Answer 2

It may be easier to delegate the increment-if-int behavior to a separate component that does just that:

template <typename T>
struct IncrementIfInt
{
  const T& operator()(const T& value) { return value; }
};

template <>
struct IncrementIfInt<int>
{
  int operator()(int value) { return value + 1; }
};

template <typename T> class foo
{
private:
  T m_var;
public:
  foo(T init_var)
    : m_var(IncrementIfInt<T>()(init_var))
  {
  }
};