RE: [Boost-Users] Why "explicit" in shared_ptr's constructor?

28 Mar 2003

      the danger comes from code like so:

void foo(boost::shared_ptr<entity> p)
{
...
}

void bar()
{
	entity* e = new entity;
	foo(e);
	delete e;
}

Allowing a pointer to be converted to a shared_ptr implicitly could cause
serious bugs.

You could shorten things with this (untested) function:

template <typename T>
boost::shared_ptr<T> make_sp(T* p)
{
	return boost::shared_ptr<T>(p);
}
...
-----Original Message-----
From: Maurizio Colucci [mailto:seguso@email.it]
Sent: Friday, March 28, 2003 8:36 AM
To: Boost-Users@yahoogroups.com
Subject: [Boost-Users] Why "explicit" in shared_ptr's constructor?
Hello,
I am a newbie user of boost.
I have a question about boost::shared_ptr. Suppose you have a
polimorphic list:
class entity{ ...};
class simple_entity : public entity{
public:
  uint id;
  simple_entity(uint i) : id(i) {}
};
int main(){
  list l;
  ...
I noticed the following code doesn't compile
l.push_back(new simple_entity(1));
too bad, because it was very easy to read, and very similar to the
behavior if standard pointers (not to speak about similarity to Java
and C#).  So I am forced to do
l.push_back(shared_ptr<entity> (new simple_entity(1)));
which is less readable and, IMHO, barely tolerable on the long
run: too much typing.
It seems that the reason is the explicit keyword in the
constructor. My question is: why? what was the danger without the
explicit?
Thank you very much for any info,
Maurizio
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RE: [Boost-Users] Why "explicit" in shared_ptr's constructor?

Tom Matelich