export function isFunction(value: any): value is (...args: any[]) => any { return typeof value === 'function'; } Why value is (...args: any[]) => any is used instead of boolean ?
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1 Answer
This is called a Type Guard which gives typescript additional type information than a runtime boolean check would. https://medium.com/swlh/typescript-generics-and-type-guards-explained-by-example-177b4a654ef6
export function isFunctionTypeGuard(value: any): value is (...args: any[]) => any { return typeof value === 'function'; } export function isFunctionBool(value: any): boolean { return typeof value === 'function'; } const doSomething = (fn: any) => { // using a runtime type check if (isFunctionBool(fn)) { fn(1, 2, 3); // ^^ typescript still thinks the type is `any` } // using a typeguard if (isFunctionTypeGuard(fn)) { fn(1, 2, 3); // ^^ typescript now knows the type is `(...args: any[]) => any` } } 
7 Comments
Wang Liang
value is (...args: any[]) => any is not returnning type of isFunction ?
xdumaine
@MERN no, that's saying "this function is asserting that the argument is a value with a type of
(...args: any[]) => any"Wang Liang
But, generally, after
:, we declared the function's return types in typescript
Roberto Zvjerković
Except with
: variable is Type syntaxxdumaine
@MarcelToth
function Foo(arg1: any): arg1 is T { ... } the arg1 is T clause indicates that the return type is boolean, and also asserts that arg1 either is or is not of type T. |