'unknown' vs. 'any'

You can read more about unknown in the PR or the RC announcement, but the gist of it is:

[..] unknown which is the type-safe counterpart of any. Anything is assignable to unknown, but unknown isn't assignable to anything but itself and any without a type assertion or a control flow based narrowing. Likewise, no operations are permitted on an unknown without first asserting or narrowing to a more specific type.

A few examples:

let vAny: any = 10;          // We can assign anything to any
let vUnknown: unknown =  10; // We can assign anything to unknown just like any 


let s1: string = vAny;     // Any is assignable to anything 
let s2: string = vUnknown; // Invalid; we can't assign vUnknown to any other type (without an explicit assertion)

vAny.method();     // Ok; anything goes with any
vUnknown.method(); // Not ok; we don't know anything about this variable

The suggested usage is:

There are often times where we want to describe the least-capable type in TypeScript. This is useful for APIs that want to signal “this can be any value, so you must perform some type of checking before you use it”. This forces users to safely introspect returned values.

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The difference between unknown and any is described as:

Much like any, any value is assignable to unknown; however, unlike any, you cannot access any properties on values with the type unknown, nor can you call/construct them. Furthermore, values of type unknown can only be assigned to unknown or any.

To answer your question of when should you use unknown over any:

This is useful for APIs that want to signal “this can be any value, so you must perform some type of checking before you use it”. This forces users to safely introspect returned values.

Take a look at the TypeScript 3.0 announcement for examples of type checking a variable of type unknown and a more detailed explanation.

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any type:

The any type represents all possible JS values. Every type is assignable to type any. Therefore the type any is an universal supertype of the type system. The TS compiler will allow any operation on values typed any. For example:

let myVar: any;

myVar[0];
myVar();
myVar.length;
new myVar();

In many occasions this is too lenient of the TS compiler. i.e. it will allow operations which we could have known to be resulting into a runtime error.

unknown type:

The unknown type represents (just like any) all possible JS values. Every type is assignable to type unknown. Therefore the type unknown is another universal supertype of the type system (alongside any). However, the TS compiler won't allow any operation on values typed unknown. Furthermore, the unknown type is only assignable to the type any. An example will clarify this:

let myVar: unknown;

let myVar1: unknown = myVar;   // No error
let myVar2: any = myVar;       // No error
let myVar3: boolean = myVar;   // Type 'unknown' is not assignable to type 'boolean'

// The following operations on myVar all give the error:
// Object is of type 'unknown'
myVar[0];
myVar();
myVar.length;
new myVar();

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any, unknown:

allow assigning any type

any:

allows being assigned to any type

allows calling any method

unknown:

doesn't allow being assigned to any type

doesn't allow calling any method

const a: any = 'a'; // OK
const b: unknown = 'b' // OK

const v1: string = a; // OK
const v2: string = b; // ERROR
const v3: string = b as string; // OK

a.trim() // OK
b.trim() // ERROR

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they are different in semantics.

unknown is the parent type of all other types. it's a regular type in the type system.

any means "turn off the type check". it's kind of meta programming.

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Unknown

If you write a function that only passes down an input to another function, use unknown. From the perspective of the function: "I don't know, I don't wanna know". There is nothing wrong with using unknown.

E.g.:

function buy(item: unknown): Purchase {
  if (item) {
    return purchase(item);
  } else {
    throw new TypeError('item is missing');
  }
}

Any

If you need to call properties on that value, then any is more suited.

Linting might not like any, suggesting you to be more specific with your input. That way, if you change the interface from isItem to isValid, typescript tells you to update your code.

E.g.:

// eslint-disable-next-line @typescript-eslint/explicit-module-boundary-types
function isItem(item: any): item is Purchase {
  return !!item?.price;
}

Calling properties

function isStuff(item: unknown): item is Stuff {
  return (item as Stuff).it !== undefined;
}
function isStuff(item: any): item is Stuff {
  return item.it !== undefined;
}
camelcaseKeys(item) as unknown as Item;

See user defined guards if you're interested, I brought it in because it's one of the few cases where I need any.

From this blog from ultimatecourses:

Use the any type when there are no other options

It's hard to find good examples for any.

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I am late to the party but will try to demystify it.

const canBeAnything: any = 100;
const canNotBeAnything: unknown = 100;

// If we try to use a .startsWith() method
canBeAnything.startsWith('10'); // no error
canNotBeAnything.startsWith('10'); // Property 'startsWith' does not exist on type 'unknown'

only way to use the method .startsWith() on unknown is explicitly telling compiler the type, like

(canNotBeAnything as string).startsWith('10'); // Chill down TS compiler, I know what I am doing. 

The latter on doesn't show any compilation error but it throws error during runtime because canNotBeAnything is a number type and we are forcing it to be string

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The accepted answer says "unknown which is the type-safe counterpart of any."

However, as this example shows unknown is its own beast and it sometimes behaves very differently from any:

type Foo = unknown extends string ? true : false // false
type Bar = any extends string ? true : false     // boolean - i.e. both true and false

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