Re:The answer here is yes!

Well, swapoff is the destruction of a complex data structure, not its construction. Its construction is the swapon system call, further down in the same file.

The same design pattern can be found all over the Linux kernel: there's a construction function, which constructs whatever complex data structure the module needs, and a destruction function, which releases it.

> If any failures occur, then all the partially-constructed pieces so far need to be freed before returning an error indication to the caller. Only if all the construction steps succeed can the complete object be returned to the caller.

In the swapon system call, there's a single error handling label "bad_swap" which frees all the partially-constructed data structures, undoes block device configuration, closes the opened file, and so on. It falls through to the "out" label, used for both the success and failure cases, which releases any temporarily used resources.

> Search through my example for the comment “so I don't dispose of it yet” to see how I deal with this. You should find three instances.

I see. You have two variables for the return value of the function: one which has its reference counter decremented at the end, and one which is actually returned from the function. You keep the allocated structure at the first one, and when everything's ready, you swap it with the second one. So in the failure case, the reference counter is decremented and it returns NULL; in the success case, the reference counter is not decremented and it returns the pointer to the structure.

It's an elegant way of doing it (though I'm annoyed at your inconsistency: twice you called it "result" and once you called it "Result"). It's also orthogonal to the use of goto versus pseudo-loop for cleanup: this trick can help simplify the code in both cases.

What you did is actually a design pattern in modern C++:

std::unique_ptr<foo> fn(/*args*/)
{
auto ret = std::make_unique<foo>(/*args*/);
/* ...code which can throw an exception... */
return ret;
}

Here, "ret" is a std::unique_ptr<foo>, which contains a pointer to a "foo". When this variable gets out of scope, which will happen if an exception is thrown, whatever is pointed to by "ret" will be deleted. When it reaches the "return ret", however, the pointer is moved (as in std::move) to the return value of the function, so when it leaves the scope, "ret" is pointing to NULL, and the returned object isn't deleted.