Can I write a C++ functor that accepts both a raw pointer and a smart pointer?

Question

Given the following:

struct Foo
{
    int bar() const;
};

struct IsEqual : public std::unary_function<Foo*, bool>
{
    int val;
    IsEqual(int v) : val(v) {}

    bool operator()(const Foo* elem) const
    {
        return elem->bar() == val;
    }
};

I have a container of Foo* and I use std::find_if and std::not1 to find out if there are any elements in the container where bar() returns something different from a given value. The code looks like this:

// Are all elements equal to '2'?
bool isAllEqual(const std::vector<Foo*> &vec)
{
    return find_if(vec.begin(), vec.end(), std::not1(IsEqual(2))) == vec.end();
}

Fast-forward into the future and I now have a different container, this time containing std::tr1::shared_ptr<Foo>. I'd love to simply re-use my functor in an overloaded version of isAllEqual(). But I can't. Foo* and shared_ptr<Foo> are different types. And I need to inherit from unary_function so I can use not1. It'd be more elegant if I could avoid writing the same functor twice.

Questions:

• Is there any way to write IsEqual so it can use both raw and smart pointers?
• Did I handcuff myself by using std::not1? Should I just write IsNotEqual instead?

Restrictions:

1. I can't use anything from the boost library.
2. Our compiler isn't cool enough to support C++0x lambdas.