Re: limiting a template class to certain types ?

"Alf P. Steinbach" <>
Thu, 13 Jul 2006 06:54:01 +0200
* Steve Hicks: wrote:

Just add a test, using the typedef T type; trick again.

template<bool b> struct test; // undefined
template<> struct test<true> { }; // defined specialization
template<typename T, bool b>
struct test_identity<T> : test<b> {
  typedef T type;

template<class T>
operator++(T x);

That's a neat trick. Unfortunately, it doesn't quite do what I need.
Here is a small example:

#include <iostream>
template <bool b> struct test;
template <> struct test<true> { };
template <class T, bool b>
struct test_identity : test<b> {
    typedef T type;
template <class T> struct is_callable {
    static const bool value = false;

template <class T> class Constant {
    T value;
    Constant(T t) : value(t) { }
    void operator()()
    { std::cout << "value: " << value << std::endl; }
template <class T> struct is_callable<Constant<T> > {
    static const bool value = true;

// T1 and T2 must be callable
template <class T1, class T2> class Sum {
    T1 t1; T2 t2;
    Sum(T1 _t1, T2 _t2) : t1(_t1), t2(_t2) { }
    void operator()() { t1(); t2(); }

template <class T1,class T2>
typename test_identity<Sum<T1,T2>,is_callable<T1>::value &&
operator+(T1 t1,T2 t2) {
    return Sum<T1,T2>(t1,t2);
/* END CODE */

Basically, it only makes sense to have Sum<T1,T2> if T1 and T2 are both
callable. I could easily insert a Boost static assert to ensure that
they are, but my issue is this: the overloaded operator+ still allows
any type to be sent to it, and then generates a compiler error if it's
not callable. So,

Constant<int> a = 5; Constant<double> b = 10.3;

works, but

Constant<int> a = 5;

fails to compile, despite my implicit constructor, since it tries to
make Sum<Constant<int>,int> and in doing so, tries to make
test<is_callable<int>::value> which is undefined.

You could try

     template <class T1,class T2>
     typename boost::enable_if_c<
         is_callable<T1>::value && is_callable<T2>::value,
     operator+(T1 t1,T2 t2) { return Sum<T1,T2>(t1,t2); }

which I believe is specifically what you're asking for.

However, for your chosen problem this leads to a little (contained)
combinatorial explosion: it answers the question but does not solve the
problem in a reasonable way, and that's because you're not asking about
the problem but about a particular non-working way to solve it.

Better to simply make sure that anything that can behave as a constant
has a value:

     #include <iostream>

     template <class T> class Constant {
         T value;
         Constant(T t) : value(t) { }
         double operator()() const
         { std::cout << "value: " << value << std::endl; return value; }

     template< typename T > T valueOf( T const& x )
     { return x; }
     template< typename T > T valueOf( Constant<T> const& x )
     { return static_cast<T>( x() ); } // Cast due to use of 'double'.

     template <class T1, class T2> class Sum {
         T1 t1; T2 t2;
         Sum(T1 _t1, T2 _t2) : t1(_t1), t2(_t2) { }
         double operator()() const { return valueOf(t1) + valueOf(t2); }

     template <class T1,class T2>
     Sum<T1,T2> operator+(T1 t1,T2 t2) { return Sum<T1,T2>(t1,t2); }

     int main()
         Constant<int> a = 5; Constant<double> b = 10.3;

         std::cout << (a+b)() << std::endl;
         std::cout << (a+10.3)() << std::endl;


A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

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