Search Terms

Discriminated union type guard destructuring destructured extracted assigned property

Suggestion

If the discriminating property of an interface is assigned to a const via destructuring or directly it is not usable as a type guard.

Use Cases

• Code brevity

Examples

In the example I would expect bool to be usable as a type guard in functions f and g but it is not. Is there something I am missing here that would make this unsound?

interface A {	bool: false; } interface B {	bool: true;	data: string; } type C = A | B; function f(x: C) {	const { bool } = x;	if (bool) { // Error: Property 'data' does not exist on type 'C'. Property 'data' does not exist on type 'A'.	return x.data;	} } function g(x: C) {	const bool = x.bool;	if (bool) { // Error: Property 'data' does not exist on type 'C'. Property 'data' does not exist on type 'A'.	return x.data;	} } function h(x: C) {	if (x.bool) {	return x.data;	} } 

https://www.typescriptlang.org/play/#src=interface%20A%20%7B%0D%0A%0D%0A%09bool%3A%20false%3B%0D%0A%7D%0D%0A%0D%0Ainterface%20B%20%7B%0D%0A%0D%0A%09bool%3A%20true%3B%0D%0A%0D%0A%09data%3A%20string%3B%0D%0A%7D%0D%0A%0D%0Atype%20C%20%3D%20A%20%7C%20B%3B%0D%0A%0D%0A%0D%0Afunction%20f(x%3A%20C)%20%7B%0D%0A%0D%0A%09const%20%7B%20bool%20%7D%20%3D%20x%3B%0D%0A%0D%0A%09if%20(bool)%20%7B%0D%0A%0D%0A%09%09return%20x.data%3B%0D%0A%09%7D%0D%0A%7D%0D%0A%0D%0A%0D%0Afunction%20g(x%3A%20C)%20%7B%0D%0A%0D%0A%09if%20(x.bool)%20%7B%0D%0A%0D%0A%09%09return%20x.data%3B%0D%0A%09%7D%0D%0A%7D

Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.