jhegedus jhegedus - 1 month ago 7
Javascript Question

sealed case classes in flow

I'm trying to emulate Scala's sealed case classes in Flow by using disjoint unions:

type ADD_TODO = {
type:'ADD_TODO',
text:string,
id:number
}

type TOGGLE_TODO = {type:'TOGGLE_TODO', id:number }

type TodoActionTy = ADD_TODO | TOGGLE_TODO


const todo = (todo:TodoTy, action:TodoActionTy) => {
switch (action.type){
case 'ADD_TODO' :
return { id:action.id, text:action.text, completed: false};
case 'TOGGGGLE_TODO': // this should give a type error
if (todo.id !== action.id) {return todo;}
return {...todo, completed:!todo.completed};
}
}


I should get a type error for
case 'TOGGGGLE_TODO':
but I don't.

Is there a way to fix that ?

EDIT:

I paste here the code from Gabriele's comment for future-proof-ness:

type TodoTy = {};

type ADD_TODO = { type: 'ADD_TODO', text: string, id: number };

type TOGGLE_TODO = { type: 'TOGGLE_TODO', id: number };

type TodoActionTy = ADD_TODO | TOGGLE_TODO;

export const todo = (todo: TodoTy, action: TodoActionTy) => {
switch (action.type){
case 'ADD_TODO': break;
// Uncomment this line to make the match exaustive and make flow typecheck
//case 'TOGGLE_TODO': break;
default: (action: empty)
}
}

Answer

The empty type can be used to verify that Flow is convinced of exhaustiveness

export const todo = (todo: TodoTy, action: TodoActionTy) => {
  switch (action.type){
    case 'ADD_TODO' :
      ...
    case 'TOGGGGLE_TODO':
      ...
    default :
      // only true if we handled all cases
      (action: empty)
      // (optional) handle return type
      throw 'unknown action'
  }
}
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