How would I achieve type inference in the following case:
type a = {
foo: number;
bar: string;
}
type b = {
foo: string;
}
let baz: a | b;
if (baz.foo === 5) {
baz.bar = "abc"; // baz type is still a | b, should be a
}
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typescriptlang.org/play/…MichaelAttard– MichaelAttard2017-06-05 11:17:23 +00:00Commented Jun 5, 2017 at 11:17
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"If we have a value that has a union type, we can only access members that are common to all types in the union." typescriptlang.org/docs/handbook/advanced-types.htmladiga– adiga2017-06-05 11:36:10 +00:00Commented Jun 5, 2017 at 11:36
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2 Answers
Apparently types cannot be inferred from the types of their properties, so you will need to define a type guard:
type a = {
foo: number;
bar: string;
}
type b = {
foo: string;
}
let baz: a | b;
function isA(x: a | b): x is a {
return typeof x.foo === 'number';
}
if (isA(baz) && baz.foo === 5) {
baz.bar = "123";
}
The isA type guard will tell TypeScript that you have checked bazs type yourself. Below is another way to achieve this with casts, but here you still need to cast baz for every usage which is probably not the best way to do this.
if ((baz as a).foo === 5) {
(baz as a).z = "123";
}
More information about type guards can be found in the TypeScript docs.
Comments
baz type is still a | b, should be a
Actually what the TypeScript compiler does is restricts property access of all unified types (a | b) to properties that exist on all of them, thus, baz.bar does not even compile because bar does not exist on type b.
