The question was about creating utility generic type which will be able to get any function type and create a new type which will have one argument more at the front.

## Solution 1: Using Parameters and ReturnType utility types

```
type AppendArgument<F extends (...args: any) => any, A>
= (x: A, ...args: Parameters<F>) => ReturnType<F>
// using
type FinalF = AppendArgument<(a: number, b: string) => number, boolean>
// FinalF is (x: boolean, a: number, b: string) => number 👍
```

What we did here:

- first argument
`F`

is narrowed to only function types`(...args: any) => any`

, such narrowing is crucial in order to use`Parameters`

and`ReturnType`

- we create a type which first argument is type
`A`

and other arguments are spread by`...`

from the original function by`Parameters`

which produce arguments types of the original type - we say that our new type returns the same thing as the original was by
`ReturnType`

## Solution 2: Using infer

```
type AppendArgument<F, A>
= F extends (...args: infer Args) => infer Return
? (x: A, ...args: Args) => Return
: never
// using
type SomeF = (a: number, b: string) => number
type FinalF = AppendArgument<SomeF, boolean>
// FinalF is (x: boolean, a: number, b: string) => number
```

The second proposition isn't using any utility types, but we use directly `infer`

which is a tool for getting types from generic types, and function is this kind of a type, parameterized by two parameters - arguments type and return type.

Notewe can understand utility types ReturnType and Parameters as types abstraction which are using`infer`

in the definition, so they are kind of abstraction over construct we did in solution 2

What we did here:

- we infer all arguments by
`...args: infer Args`

and return type by`=> infer Return`

- we are putting newly obtained types in the definition, with putting as first argument type
`A`

exactly as we did in solution 1 - take a look that we needed to use conditional type, as this is the only way we can work with
`infer`

If you want to know more about `infer`

check the Answer 1 from the series

**This series is just starting**. If you want to know about new exciting questions from advanced TypeScript please follow me on dev.to and twitter.

## Discussion

I would like to use

`unknown`

over`any`

in`AppendArgument`

. Because I would prefer not to get past type checking using`any`

.Any ideas of how I can change this?

Extends any[] doesn't mean it is any[] it only means it is any array, so array with any type you want. Compiler will infer correct type and it will never be any. There is no type whole here