[C++] Fix compilation with enable_shared_from_this

Summary:
Fixes issue with template argument deduction with enable_shared_from_this as argument
```
#include<memory>

template<class T>
void makeWeak(const std::shared_ptr<T>& x) {}

struct X : public std::enable_shared_from_this<X>{
};

void test() {
  X x
  makeWeak(x.shared_from_this()); // compilation failed here - it was unable to deduce template parameter of makeWeak
}
```

Reviewed By: jvillard

Differential Revision: D4414788

fbshipit-source-id: 4d19c53

infer/models/cpp/include/infer_model/begin_name_override.inc

@@ -1,4 +1,5 @@
 #define make_shared std__make_shared
+#define enable_shared_from_this std__enable_shared_from_this
 #define shared_ptr std__shared_ptr
 #define unique_ptr std__unique_ptr
 #define make_unique std__make_unique

infer/models/cpp/include/infer_model/end_name_override.inc

@@ -1,4 +1,5 @@
 #undef make_shared
+#undef enable_shared_from_this
 #undef shared_ptr
 #undef unique_ptr
 #undef make_unique

infer/models/cpp/include/infer_model/shared_ptr.h

@@ -392,6 +392,17 @@ shared_ptr<T> const_pointer_cast(shared_ptr<U> const& r) noexcept {
   return const_pointer_cast<T, U>((const std__shared_ptr<U>&)r);
 }
 
+template <class T>
+class enable_shared_from_this : public std__enable_shared_from_this<T> {
+ public:
+  shared_ptr<T> shared_from_this() {
+    return std__enable_shared_from_this<T>::shared_from_this();
+  }
+  shared_ptr<T const> shared_from_this() const {
+    return std__enable_shared_from_this<T>::shared_from_this();
+  }
+};
+
 template <class T, class... Args>
 shared_ptr<T> make_shared(Args&&... args) {
   return shared_ptr<T>(new T(std::forward<Args>(args)...));