C++ – getting a normal ptr from shared_ptr


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?

Best Solution

Use the get() method:

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

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.