RE: [boost] Always explicit , why?

From: Vugts, Robert (robert.vugts_at_[hidden])
Date: 2000-09-04 09:27:15

Whilst we are on the subject: Why is the constructor
that accepts an std::auto_ptr not explicit ?

    -- Rob

-----Original Message-----
From: Gernot Neppert [mailto:gn_at_[hidden]]
Sent: 04 September 2000 15:12
To: boost_at_[hidden]
Subject: [boost] Always explicit , why?

I've been using a reference-counted pointer class extensively in my 
c++ programs, and I'd like to switch completely to boost::shared_ptr.
But there's one thing that puts me off: The constructor that accepts 
a raw pointer is declared "explicit".
When you're passing around pointers as shared_ptr parameters, it's 
just unnerving always having to provide the explicit ctor-call.
I couldn't come up with a compelling reason why this should be 
necessary. When there's a function that expects a shared_ptr<Type>, 
why not provide it simply a "new Type()"? 
Am I overlooking any ambiguities or unexpected behavour that could 
result from removing the "explicit" from shared_ptr's ctor?
Or is it plainly considered heresy to deviate from Standard 
Library's "auto_ptr" pattern ;-)

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