Suppose I want to enforce the limitation that the template parameter passed in is a child of Foo.
Is there a way to enforce this through type-traits?  A compile-time static_assert failure would be great.
In the code below, let's make it a two-part (separate) question.
My_Limited_Template<Bar> to compile.My_Limited_Template<TBar> to compile.EDIT
I apologize for the bad naming: TBar and TBaz are meant to be non-template classes on purposes.  I just attached T in front of the names to disambiguate from the classes in Part 1.
CODE
struct Foo { };                // no
struct Bar : public Foo { };   // yes
struct Baz { };                // no
template< typename T >
struct TFoo { };                       // no
struct TBar : public TFoo<TBar> { };   // yes
struct TBaz { };                       // no
template< typename T >
struct My_Limited_Template
{
  // Part One:
  //   My_Limited_Template<Foo>  // disallow
  //   My_Limited_Template<Bar>  // allow
  //   My_Limited_Template<Baz>  // disallow
  // 
  // Part Two:
  //   My_Limited_Template<TFoo<int>> // disallow
  //   My_Limited_Template<TBar>      // allow
  //   My_Limited_Template<TBaz>      // disallow
};
I assume that you made an error in definition of TBar and TBas, check that my modification is correct. 
#include <type_traits>    
struct Foo { };                // don't allow this
struct Bar : public Foo { };   // allow this
struct Baz { };                // don't allow this
template< typename T > struct TFoo { };                       
template< typename T > struct TBar : public TFoo<TBar<T>> { }; 
template< typename T > struct TBaz { };                       
template< typename T >
struct My_Limited_Template
{
        static_assert(
                (std::is_base_of<Foo,T>::value && !std::is_same<T,Foo>::value)
                || 
                (std::is_base_of<TFoo<T>,T>::value && !std::is_same<T,TFoo<T>>::value),
                "fail2"
        ); 
};
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