如何为具有特定类型特征的所有类型编写函数模板？

Question

Consider the following example:

struct Scanner
{
    template <typename T>
    T get();
};

template <>
string Scanner::get()
{
    return string("string");
}

template <>
int Scanner::get()
{
    return 10;
}

int main()
{
    Scanner scanner;
    string s = scanner.get<string>();
    int i = scanner.get<int>();
}

The Scanner class is used to extract tokens from some source. The above code works fine, but fails when I try to get other integral types like a char or an unsigned int . The code to read these types is exactly the same as the code to read an int . I could just duplicate the code for all other integral types I'd like to read, but I'd rather define one function template for all integral types.

I've tried the following:

struct Scanner
{
    template <typename T>
    typename enable_if<boost::is_integral<T>, T>::type get();
};

Which works like a charm, but I am unsure how to get Scanner::get<string>() to function again. So, how can I write code so that I can do scanner.get<string>() and scanner.get<any integral type>() and have a single definition to read all integral types?

Update: bonus question : What if I want to accept more than one range of classes based on some traits? For example: how should I approach this problem if I want to have three get functions that accept (i) integral types (ii) floating point types (iii) strings, respectively.

Answer 1

struct Scanner
{
    template <typename T>
    typename boost::enable_if<boost::is_integral<T>, T>::type get()
    {
        return 10;
    }
    template <typename T>
    typename boost::disable_if<boost::is_integral<T>, std::string>::type get()
    {
        return "string";
    }
};

Update "What if I want to accept more than one range of classes based on some traits?"

struct Scanner
{
    template <typename T>
    typename boost::enable_if<boost::is_integral<T>, T>::type get()
    {
        return 10;
    }

    template <typename T>
    typename boost::enable_if<boost::is_floating_point<T>, T>::type get()
    {
        return 11.5;
    }

    template <typename T>
    std::string get(
          typename boost::disable_if<boost::is_floating_point<T>, T>::type* = 0, 
          typename boost::disable_if<boost::is_integral<T>, T>::type* = 0)

    {
        return std::string("string");
    }
};

Answer 2

Defer to another template. Here's the general pattern for what you want:

template <typename T, bool HasTrait = false>
struct scanner_impl;

template <typename T>
struct scanner_impl
{
    // Implement as though the trait is false
};

template <typename T>
struct scanner_impl<true>
{
    // Implement as though the trait is true
};

// This is the one the user uses
template <typename T>
struct scanner : scanner_impl<T, typename has_my_trait<T>::value>
{
};