Assert functions do not constraint type when they are guarded by a truthy expression.

[mhevery]

TypeScript Version: 3.8.3

Search Terms:: Assert Function

Code

function assertString(actual: any): asserts actual is string {}
function expectString(value: string) { }
let devMode: {} = {};

// This works as expected
const x: string|number = '' as any;
assertString(x);
expectString(x); // x is string!

// This fails
const y: string | number = '' as any;
// Expecting TS to correctly infer that `devMode` is always truthy and 
// hence the`assertString` sholud always execute therefore its 
// type assertion should hold.
devMode && assertString(y);
expectString(y); // y is string|number hence it fails

Expected behavior:

Assert functions should be assumed to always execute when they are guarded by an expression which is truthy.

Actual behavior:

Any sort of guard expression (even if know to be truthy) prevents the assert function from applying its constraint.

Playground Link:
Playground

Related Issues:

• Assertion functions cannot be used to narrow discriminated union types · Issue #37241 · microsoft/TypeScript
• Assertion functions don't work when defined as methods · Issue #36931 · microsoft/TypeScript

[RyanCavanaugh]

We don't trust the results of our own truthiness analysis for these situations because there are legitimate situations where a falsy value might inhabit a truthy position. We also believe that code like x && y is written because x is possibly falsy (otherwise why would you do it...?). It seems like you intend to strip this out later, so I'd recommend just using some other approach.

[typescript-bot]

This issue has been marked 'Working as Intended' and has seen no recent activity. It has been automatically closed for house-keeping purposes.

[typescript-bot]

This issue has been marked 'Working as Intended' and has seen no recent activity. It has been automatically closed for house-keeping purposes.