Conditional Types In Flow

Is it possible to type a variable in flow based on a condition? Something like this: const type = 'xyz'; const a: (type === 'xyz') ? number : string;

Solution 1:

Type-level conditions in Flow may be simulated using type calls ($Call type):

type $If<X: boolean, Then, Else = empty> = $Call<
    & ((true, Then, Else) => Then)
    & ((false, Then, Else) => Else),
    X,
    Then,
    Else,
>;

type $Not<X: boolean> = $If<X, false, true>;
type $And<X: boolean, Y: boolean> = $If<X, Y, false>;
type $Or<X: boolean, Y: boolean> = $If<X, true, Y>;

type $Gte<X, Y> = $Call<
    & ($Subtype<X> => true)
    & (mixed => false),
    Y,
>;

// Usage example:

declare var a: $Gte<number, string>;

/* error  1 */ (a: true);
/* ok       */ (a: false);

declare var b: $Gte<number, number>;

/* ok       */ (b: true);
/* error  2 */ (b: false);

declare var c: $If<true, 1, 2>;

/* ok       */ (c: 1);
/* error  3 */ (c: 2);

declare var d: $If<false, 1, 2>;

/* error  4 */ (d: 1);
/* ok       */ (d: 2);

More usage examples may be found in gist.

Solution 2:

The short answer is no. A variable must be bound to a single type, which can include a union type like number | string. Inside a conditional block Flow can infer a more specialised type. For example:

const a: number | string = "foo";
if(typeof a === "string"){
  a.split(); // OK!
} else {
  a.split(); // Error! a is a number
}

But notice that these conditional branches are static, and it's guaranteed at compile time what the type will be inside each.