C++ 从shared_ptr 得到一个普通的ptr？
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getting a normal ptr from shared_ptr?
提问by Adam Rosenfield
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?
我有类似的东西，shared_ptr<Type> t(makeSomething(), mem_fun(&Type::deleteMe))我现在需要调用需要指向Type. 我如何获得它shared_ptr？
回答by Adam Rosenfield
Use the get()method:
使用get()方法：
boost::shared_ptr<foo> foo_ptr(new foo());
foo *raw_foo = foo_ptr.get();
c_library_function(raw_foo);
Make sure that your shared_ptrdoesn'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.
确保shared_ptr在库函数完成之前您没有超出范围 - 否则可能会导致糟糕的结果，因为库可能会在指针被删除后尝试对其进行处理。如果库函数在返回后维护原始指针的副本，则要特别小心。
回答by Michael Burr
Another way to do it would be to use a combination of the &and *operators:
另一种方法是使用&和*运算符的组合：
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.
虽然我个人更喜欢使用该get()方法（它确实是正确的答案），但它的一个优点是它可以与重载operator*（指针取消引用）但不提供get()方法的其他类一起使用。例如，在泛型类模板中可能很有用。
