getting a normal ptr from shared_ptr?

Question

I have something like shared_ptr<Type> t(makeSomething(), mem_fun(&Type::deleteMe)) I now need to call C styled function that requires a pointer to Type. How do I get it from shared_ptr?

Answer 1

Use the get() method:

boost::shared_ptr<foo> foo_ptr(new foo());
foo *raw_foo = foo_ptr.get();
c_library_function(raw_foo);

Make sure that your shared_ptr doesn't go out of scope before the library function is done with it -- otherwise badness could result, since the library may try to do something with the pointer after it's been deleted. Be especially careful if the library function maintains a copy of the raw pointer after it returns.

Answer 2

Another way to do it would be to use a combination of the & and * operators:

boost::shared_ptr<foo> foo_ptr(new foo());
c_library_function( &*foo_ptr);

While personally I'd prefer to use the get() method (it's really the right answer), one advantage that this has is that it can be used with other classes that overload operator* (pointer dereference), but do not provide a get() method. Might be useful in generic class template, for example.

Answer 3

For std::shared_ptr (C++11 onwards) also, there is a get method to obtain the raw pointer. link