How do you represent this in a way that makes it impossible, even through programmer error elsewhere in the program, to have the flags in an invalid state?
You can create en enum that looks like:
enum program_state =
(
W_X_Y_NZ,
W_NX_NY_Z,
NW_NX_NY_Z,
... and so on
); type ConstrainedByW =
| { w: false, x: false }
| { w: true, x: bool }
type ConstrainedByY =
| { x: bool, y: false, z: bool }
| { x: true, y: true, z: false }
type ProgramState = ConstrainedByW & ConstrainedByYComplexity is both subjective and relative. Allowing invalid states in your model is going to complicate things, too.
I wouldn't mind having this example in production. It will scale just fine with the number of booleans.
enum program_state {
X_W,
Y_X_W,
Z,
W,
Z_W,
Z_X_W,
}
You only have these 6 options. And in this case you can easily use a SAT solver to generate the cases for you so that you don't have to write them by hand.Rather than thinking of it as an enum, think of it as a list of contructors:
class ProgramState {
bool w, x, y, z;
ProgramState(x, z) // implies y = true, w = true
ProgramState(w, z) // cannot set x; implies y = false (cannot set y)
}
Even if the class needs all four fields, internally to represent all the possible combinations of data, there's no constructors/setters to work with them independently. (Which is also related to why "Make Impossible States Unrepresentable" also generally implies immutable, no setters at all makes it much easier to make states impossible.)In languages with discriminated unions you might even have some natural field sharing depending on how your "constructors" are written and the memory expansion isn't necessarily "exponential".
// idiomatic typescript
type Optional<IfAbsent, IfPresent> =
| {type: 'absent', detail: IfAbsent}
| {type: 'present', value: IfPresent}
// Church-encoded version
type Either<x, y> = <z>(ifLeft: (x: x) => z, ifRight: (y: y) => z) => z
// isomorphism between the two
function church<x, y>(opt: Optional<x, y>): Either<x, y> {
return (ifLeft, ifRight) => opt.type === 'absent'? ifLeft(opt.detail) : ifRight(opt.value)
}
function unchurch<x, y>(opt: Either<x, y>): Optional<x, y> {
return opt<Optional<x,y>>(x => ({type: 'absent', detail: x}), y => ({type: 'present', value: y}))
}
In addition the Church encoding of a sum type, is a function that takes N handler functions and calls the appropriate one for the case that the data type is in. With a little squinting, this is the Visitor pattern. interface LeftRightVisitor<X, Y, Z> {
visit(x: Left<X>): Z
visit(y: Right<Y>): Z
}
interface LeftRight<X, Y> {
accept<Z>(visitor: LeftRightVisitor<X, Y, Z>): Z;
}
class Left<X> implements LeftRight<X, any> {
constructor(public readonly x: X) {}
accept<Z>(visitor: LeftRightVisitor<X, any, Z>) {
return visitor.visit(this)
}
}
class Right<Y> implements LeftRight<any, Y> {
constructor(public readonly y: Y) {}
accept<Z>(visitor: LeftRightVisitor<any, Y, Z>) {
return visitor.visit(this)
}
}
// isomorphism
function visitify<X, Y>(opt: Optional<X, Y>): LeftRight<X, Y> {
return opt.type === 'absent' ? new Left(opt.detail) : new Right(opt.value)
}
function unvisitify<X, Y>(opt: LeftRight<X, Y>): Optional<X, Y> {
return opt.accept({
visit(value: Left<X> | Right<Y>) {
return value instanceof Left? {type: 'absent', detail: value.x} : {type: 'present', value: value.y}
}
})
}
The main difference with the usual visitor pattern is that the usual visitor pattern doesn't return anything (it expects you to be holding some mutable state and the visitor will mutate it), you can do that too if you don't have access to a suitable generic for the Z parameter.