Why doesn't auto_ptr<T> have operator!() defined?

Question

The Title pretty much sums up my question. Why can't the following be done to check for a null pointer?

auto_ptr<char> p( some_expression ); // ... if ( !p )  // error 

This must be done instead:

if ( !p.get() ) // OK 

Why doesn't auto_ptr<T> simply have operator!() defined?

Answer 1

Seems to be there was an error in its design. This will be fixed in C++0x. unique_ptr (replacement for auto_ptr) contains explicit operator bool() const;

Quote from new C++ Standard:

The class template auto_ptr is deprecated. [Note: The class template unique_ptr (20.9.10) provides a better solution. —end note ]

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Some clarification:
Q: What's wrong with a.get() == 0?
A: Nothing is wrong with a.get()==0, but smart pointers lets you work with them as they were real pointers. Additional operator bool() gives you such a choice. I think, that the real reason for making auto_ptr deprecated is that is has has not intuitive design. But operator bool for unique_ptr in the new Standard means that there are no reasons not to have it.