Give a more helpful error message when Gen is constructed with incompatible type

This produces a compile error in Gen's constructor.  Before this change,
errors would be attributed to GenImpl.

Author: cirodrig

include/rapidcheck/Gen.hpp

@@ -1,5 +1,6 @@
 #pragma once
 
+#include <type_traits>
 #include <cassert>
 
 #include "rapidcheck/detail/Any.h"
@@ -17,6 +18,13 @@ Gen<Decay<typename std::result_of<Mapper(T)>::type>> map(Gen<T> gen,
 
 } // namespace gen
 
+// Concept check for types that have method
+// `Shrinkable<T> G::operator()(const Random &, int) const`
+template<typename T, typename G>
+using MakesShrinkable = std::is_convertible<
+  decltype(std::declval<const G>()(std::declval<Random>(), 0)),
+  Shrinkable<T>>;
+
 template <typename T>
 class Gen<T>::IGenImpl {
 public:
@@ -55,7 +63,10 @@ class Gen<T>::GenImpl : public IGenImpl {
 template <typename T>
 template <typename Impl, typename>
 Gen<T>::Gen(Impl &&impl)
-    : m_impl(new GenImpl<Decay<Impl>>(std::forward<Impl>(impl))) {}
+    : m_impl(new GenImpl<Decay<Impl>>(std::forward<Impl>(impl))) {
+  static_assert(MakesShrinkable<T, Decay<Impl>>::value,
+                "Generator implementation must have a method Shrinkable<T> operator()(const Random &, int) const");
+}
 
 template <typename T>
 std::string Gen<T>::name() const {