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Gilbert for The Clause

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Write a User Permissions System in 5 lines of Prolog

Logic programming is something all developers should learn. It isn't actually that hard, and simply knowing what's possible will positively influence you in choosing better tools, for now and the future.

In this post I'm going to show you how easy it is to write a role-based permissions system in Prolog. In our first iteration, you'll only need 5 lines of logic, and easily extendable to boot. This will give you a sample taste of the full power of Prolog.

Note: This is not a full introduction to Prolog. That said, you should still be able to follow along, because logic programming is easy. When you're ready to dive deeper, you can bookmark this post for later.

The Scenario

Let's say we have users and clubs:


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with a many-to-many relationship between the two:

member(alice, chess).
member(dan,   chess).
member(elli,  chess).

member(alice, boxing).
member(bob,   boxing).
member(carly, boxing).
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And lastly, we want to specify which user is allowed to moderate which club:

moderator(dan,   chess).
moderator(alice, boxing).
moderator(bob,   boxing).
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Note that we haven't yet written a single line of logic. All code so far has been facts – simple statements of truth. And yet, with only this, Prolog grants us great, expressive power! Allow me to demonstrate.

Here is how we ask Prolog for all the clubs alice is a member of:

?- member(alice, C).
C = chess ;
C = boxing.
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Note how Prolog gives us two solutions.

How about all clubs that alice is also a moderator of?

?- member(alice, C), moderator(alice, C).
C = boxing.
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Or how about all non-moderator members of all clubs?

?- member(U, C), \+ moderator(U, C).
U = alice,
C = chess ;
U = elli,
C = chess ;
U = carly,
C = boxing.
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Note how Prolog gives us three solutions when there are six memberships. Also note that \+ is the operator for "not".

The Logic

Given the incredible expressiveness Prolog grants us, it's very easy to begin adding logic for protected moderator actions.

Let's say we want a moderator to be able to ban a club member. A little harsh, I know, but it's a concept we all understand!

Here's how you can do it – in the advertised 5 lines – to be explained shortly afterwards:

can(User, Club, ban_user, Target) :-
  moderator(User, Club),
  dif(User, Target),
  member(Target, Club),
  \+ moderator(Target, Club).
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and here's how you use it:

% Alice can ban a member
?- can(alice, boxing, ban_user, carly).

% But she cannot ban another moderator
?- can(alice, boxing, ban_user, bob).

% Nor can she ban herself
?- can(alice, boxing, ban_user, alice).

% And non-moderators cannot ban anyone
?- can(elli, chess, ban_user, alice).
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That's right, we now have fully working permission logic for banning a user. All moderators now have the ability to ban. To add another moderator, just add a new moderator clause.

But wait, there's more. We can still ask questions! Who can Dan ban?

?- can(dan, Club, ban_user, Person).
Club = chess,
Person = alice ;
Club = chess,
Person = elli.
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Prolog gives us all the answers. Dan has the permission to ban Alice and Elli from the chess club. Wow!

Code Explanation

Here's an explanation of the can rule:

  • can(User, Club, ban_user, Target) - For a given user, club, and target, we are defining logic specifically for ban_user.
  • moderator(User, Club) - The given user must be a moderator of the given club.
  • dif(User, Target) - And lastly, the user performing the action cannot target themselves (dif is built into Prolog).
  • member(Target, Club) - The given target user must be a member of that same club. You can't ban someone who's not even in the club!
  • \+ moderator(Target, Club) - The target user must not be a moderator.

And that's all it takes. Gaze upon the beauty of Prolog.

You can see the full code here.

Going Further

  • The facts we wrote are hard-coded. In practice, these can be loaded from a database.
  • The concept of roles and permissions can be decoupled and abstracted further, making our data even more serialization-friendly. This is covered in Part 2.
  • SWI Prolog has HTTP and Docker support, so it's possible to host a "logic server" that answers questions for other web services you control.

Interested in learning more about Prolog? Star our repo, join our chatroom, and support our Patreon!

See Part 2 of this series here:

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