Typescript strict check type in conditional types

Question

In the following code I want type A to be true, and not the union of both:

type isFalseAndNotAny<T> = T extends false ? 'false' : 'true';

// how do I get here only `true`?
type A = isFalseAndNotAny<any>; 
    
// that's OK we get only `false`    
type B = isFalseAndNotAny<false>;

Is there any way in Typescript to do something similar like in Javascript we use the triple equality === operator?

score 1 · Accepted Answer · answered Sep 14 '21 at 07:19

You need slightly modify your util.

// credits goes to https://stackoverflow.com/questions/55541275/typescript-check-for-the-any-type
type IfAny<T, Y, N> = 0 extends (1 & T) ? Y : N;
type IsAny<T> = IfAny<T, true, false>;

type isFalseAndNotAny<T> = T extends false ? IsAny<T> extends false ? 'false' : 'true' : 'true';

// true
type A = isFalseAndNotAny<any>;

// false
type B = isFalseAndNotAny<false>;

You need explicitly check if T extends false and not any, it is like T===false && T!==any

Typescript strict check type in conditional types

1 Answers1