858

Code is:

const foo = (foo: string) => {
  const result = []
  result.push(foo)
}

I get the following TS error:

[ts] Argument of type 'string' is not assignable to parameter of type 'never'.

What am I doing wrong? Is this a bug?

1
  • 9
    const result: [] = [] => const result: string[], not just [], but string[], if you don't know the type inside the array, you can always fallback to any[]
    – jave.web
    Commented Apr 9, 2022 at 15:37

21 Answers 21

1107

All you have to do is define your result as a string array, like the following:

const result : string[] = [];

Without defining the array type, it by default will be never. So when you tried to add a string to it, it was a type mismatch, and so it threw the error you saw.

13
  • 74
    That seems bizarre to me, why default to never[] ? is there ever a use for that except "an array that will ALWAYS be empty?" Commented Feb 28, 2019 at 20:56
  • 16
    I totally agree because, to be honest, I can't think of a use case that an array of nevers will be useful. But while looking into it I came across this answer which has great info. Commented Mar 1, 2019 at 11:42
  • 45
    Shouldn't the default type be "any"? When you declare a variable (let x; ) it is of type any by default. Not never. Commented May 7, 2019 at 8:53
  • 10
    Definitely would expect an array without type to be any[]. Was surprised it was never[] which makes it useless. Unintuitive. Thanks for the answer. edit: grammar
    – Igor Malyk
    Commented Dec 24, 2019 at 10:38
  • 20
    @VincentBuscarello I guess, the main point of such default is to make one to always add types to their arrays; but the error message is definitely not helpful.
    – YakovL
    Commented Dec 27, 2019 at 14:24
138

Another way is:

const result: any[] = [];
4
  • 88
    Actually using any isn't a good option, because it basically turns off any TypeScript features. It's a workaround and can lead to various bugs. Commented Jul 12, 2019 at 9:05
  • 11
    Yes. But sometimes nothing else works especially when you are working with third-party libraries..
    – neomib
    Commented Jul 14, 2019 at 7:53
  • 18
    wouldnt const result: any[] = [] be more accurate? its an array of any not and array treated as any
    – g00dnatur3
    Commented Jan 6, 2021 at 23:07
  • 1
    @g00dnatur3 is right. If you must use any, then at least specify that it is an array of items of any type.
    – devaent
    Commented Nov 23, 2021 at 16:16
129

This seems to be some strange behavior in typescript that they are stuck with for legacy reasons. If you have the code:

const result = []

Usually it would be treated as if you wrote:

const result:any[] = []

however, if you have both noImplicitAny FALSE, AND strictNullChecks TRUE in your tsconfig, it is treated as:

const result:never[] = []

This behavior defies all logic, IMHO. Turning on null checks changes the entry types of an array?? And then turning on noImplicitAny actually restores the use of any without any warnings??

It's unintuitive that adding an additional type check (noImplicitAny) results in more relaxed type checking, especially when it introduces the thing you were attempting to eliminate.

4
  • 9
    True, this is absolutely unintuitive, a parameter called "no implicit any" should imply that any time something would be assumed to be any - it is not. Also, the implicit "never" seems dumb, why not simply state an error with: Array does not have its type declared. IMHO this should solely depend on noImplicitAny, not on the combination.
    – CodeSmith
    Commented Feb 17, 2021 at 9:49
  • 2
    I think that people who contribute to large repos tend to sometimes over-engineer it. At least we know what flags we should set to make it intuitive for us. As for me I set both flags to true and feel fine with it.
    – fires3as0n
    Commented Feb 22, 2021 at 15:16
  • Just to clarify, what is changed is not the runtime behavior of the program (there are no types at runtime!). The only thing that changes is the type checker behavior, which becomes a little bit more strict as you enable more checking options in your tsconfig. I don't think it "defies all logic"
    – Brann
    Commented Mar 31, 2023 at 4:45
  • @Brann Not quite, it becomes LESS strict as you enable the "no implicit any" checking options. Commented Jan 24 at 15:05
123

I got the same error in React function component, using useState hook.

The solution was to declare the type of useState at initialisation using angle brackets:

// Example: type of useState is an array of string
const [items , setItems] = useState<string[]>([]); 
0
36

I was having same error In ReactJS statless function while using ReactJs Hook useState. I wanted to set state of an object array , so if I use the following way

const [items , setItems] = useState([]);

and update the state like this:

 const item = { id : new Date().getTime() , text : 'New Text' };
 setItems([ item , ...items ]);

I was getting error:

Argument of type '{ id: number; text: any }' is not assignable to parameter of type 'never'

but if do it like this,

const [items , setItems] = useState([{}]);

Error is gone but there is an item at 0 index which don't have any data(don't want that).

so the solution I found is:

const [items , setItems] = useState([] as any);
2
  • 4
    Actually, I think a better version of your solution would be const [items , setItems] = useState([] as string[]); Commented Jan 29, 2021 at 12:09
  • A better approach would be to declare the type of useState at the time of initialization, as suggested by @neiya: const [items , setItems] = useState<any[]>([]); Commented Oct 26, 2023 at 4:12
32
const foo = (foo: string) => {
  const result: string[] = []
  result.push(foo)
}

You needed specify what the array is since result = [] has a return type of any[]. Typically you want to avoid any types since they are meant to be used as an "Escape hatch" according to Microsoft.

The result is an object that is an array of string.

25

The solution i found was

const [files, setFiles] = useState([] as any);
5
  • 9
    Using as any thwarts a static typing system. If you know what type or types will be housed in the array, it's much better to be explicit: const myArray: string[] = []; const [files, setFiles] = useState(myArray);. Note in the case of useState, you can pass in a generic, instead: const [files, setFiles] = useState<string[]>([]);
    – jmealy
    Commented Oct 1, 2020 at 20:00
  • This is a bad practice. By taking as any you are just saying "nah I don't want to use typescript". Neiya's answer on the other hand is good enough. A better approach should be interface, but I am new to typescript as well, so I am not sure how to interface this one.
    – Gogol
    Commented Jul 3, 2021 at 9:20
  • 3
    I can't believe this answer has so many upvotes, this is basically silencing typescript, defeating its purpose altogether.
    – jperl
    Commented Oct 4, 2021 at 14:14
  • 1
    @jmealy your answer worked for me and should be the ts way of solving this. Following your example, I declared a const outside my component as const initialEmailsState: string[] = []; and used it inside the the component with useState as const [emails, setEmails] = useState(initialEmailsState); and now I can set emails like this without problems: setEmails([...emails, createOption(inputValue)]);
    – Chaiwa
    Commented Jul 20, 2022 at 19:49
  • dont do this. please use generic for element typing.
    – ZeroCho
    Commented Jul 30, 2022 at 1:36
12

I was able to get past this by using the Array keyword instead of empty brackets:

const enhancers: Array<any> = [];

Use:

if (typeof devToolsExtension === 'function') {
  enhancers.push(devToolsExtension())
}
2
  • Thanks for pointing out. The Array keyword yet is another good way. It is even more semantically useful in case you have to define a type for an array of arrays like this let parsed: Array<[]> = []; Commented Oct 23, 2020 at 8:39
  • @ValentineShi You could still do something like [][]. If you wanted something like an array of array of numbers you could do number[][]. On that note, do consider that: parsed: Array<[]> here is implicitly Array<any[]>
    – Shah
    Commented Mar 24, 2022 at 15:43
12

Error: Argument of type 'any' is not assignable to parameter of type 'never'.

In tsconfig.json -

  "noImplicitReturns": false,

   "strictNullChecks":false,

enter image description here

Solution: type as 'never'

enter image description here

1
  • Horrible solution. Can't believe that this has been upvoted so much. Use proper types instead, or create them by yourself. Do not use type any. Commented Mar 4 at 12:41
6

Remove "strictNullChecks": true from "compilerOptions" or set it to false in the tsconfig.json file of your Ng app. These errors will go away like anything and your app would compile successfully.

Disclaimer: This is just a workaround. This error appears only when the null checks are not handled properly which in any case is not a good way to get things done.

1
  • For me,in tsconfig is only "strict" part,when i set it to false as you say error go away,and this is in react,not angular.Thanks. Commented Aug 16, 2020 at 8:45
6

One more reason for the error.

if you are exporting after wrapping component with connect()() then props may give typescript error
Solution: I didn't explore much as I had the option of replacing connect function with useSelector hook
for example

/* Comp.tsx */
interface IComp {
 a: number
}

const Comp = ({a}:IComp) => <div>{a}</div>

/* ** 

below line is culprit, you are exporting default the return 
value of Connect and there is no types added to that return
value of that connect()(Comp) 

** */

export default connect()(Comp)


--
/* App.tsx */
const App = () => {
/**  below line gives same error 
[ts] Argument of type 'number' is not assignable to 
parameter of type 'never' */
 return <Comp a={3} />
}
6

I got the error when defining (initialising) an array as follows:

let mainMenu: menuObjectInterface[] | [] = [];

The code I got the problem in:

let mainMenu: menuObjectInterface[] | [] = [];
dbresult.rows.forEach((m) => {
    if (!mainMenu.find((e) => e.menucode === m.menucode)) {
        // Not found in mainMenu, yet
        mainMenu.push({menucode: m.menucode, menudescription: m.menudescription})  // Here the error
    }
})

The error was: TS2322: Type 'any' is not assignable to type 'never'

The reason was that the array was initialised with also the option of an empty array. Typescript saw a push to a type which also can be empty. Hence the error.

Changing the line to this fixed the error:

let mainMenu: menuObjectInterface[] = [];
4

You need to type result to an array of string const result: string[] = [];.

3

I did a workaround by doing

const array = new Array()
array.push(123)
console.log(array) // => [123]
1
  • 1
    It stopped the error on the type, I understand its better to declare the types, but in some extraordinary case, I want some javascript flexibility...
    – Mark
    Commented Jul 25, 2023 at 14:09
3

In case you are using TypeScript with Vue and you are facing this issue in data when initializing the variable. you can try this {} as any.

export default defineComponent({
name: "Student",
 data() {
    return {
       studentList: {} as any,
    }
}
......
})

Note: Usually as a best practice, we discourage any, so replace any with your Object model. like studentList: {} as StudentModel. where Student model can be a Interface like this:

export interface StudentModel {
    studentId: number,
    fullName: string,
    grade: string
}
2
  • 1
    Not to beat a dead horse, but use of any should be discouraged in favour of using TS properly. I use it too here and there, however it's just a convenient escape hatch. This is in case a newbie glances at this answer.
    – Kalnode
    Commented Oct 19, 2023 at 14:42
  • yeah makes sense @kalnode, just in place of any, we can replace with the Object model that we have. let me add this in answer as well.
    – KushalSeth
    Commented Oct 20, 2023 at 6:56
2

you could also add as string[]

const foo = (foo: string) => {
  const result = []
  (result as string[]).push(foo)
}

I did it when it was part of an object

let complexObj = {
arrData : [],
anotherKey: anotherValue
...
}
(arrData as string[]).push('text')
2

In Vue 3 app for typescript supporting , you can define as follows :

const result = ref<string[]>([]);

const foo = (foo:string)=>{
    result.value.push(foo as string)
}
1

I received the same error

Argument of type 'string' is not assignable to parameter of type 'never'.

in a jasmine test:

const toggleModalSpy = spyOn(component, 'toggleUpload').and.callThrough();

The actual error was that the component had no 'toggleUpload' method. It took me some time to figure it out. 😂

0
0

in latest versions of angular, you have to define the type of the variables:

  1. if it is a string, you must do like that:

    public message : string ="";

  2. if it is a number:

    public n : number=0;

  3. if a table of string:

    public tab: string[] = [];

  4. if a table of number:

    public tab: number[]=[];

  5. if a mixed table:

    public tab: any[] = []; .......etc (for other type of variables)

  6. if you don't define type of variable: by default the type is never

NB: in your case, you must know the type of variables that your table must contain, and choose the right option (like option 3 ,4 5 ).

1
  • 2
    This is not related to Angular, it's a typescript question
    – José Neto
    Commented Sep 2, 2022 at 11:03
0

This error occurs when you set the value of a property in an object when you haven't set the properties of the object. Declare a type for the object with its properties, then assign the type when you instantiated the object. Now you can set the values of the properties without this error.

0

I had this error for useRef

before with error:

const revealRefs = useRef([]);

  const addToRefs = (el) => {
    if (el && !revealRefs.current.includes(el)) {
      revealRefs.current.push(el);
    }
  };

after without error:

const revealRefs = useRef<HTMLElement[]>([]);

  const addToRefs = (el: HTMLElement | null) => {
    if (el && !revealRefs.current.includes(el)) {
      revealRefs.current.push(el);
    }
  };

Not the answer you're looking for? Browse other questions tagged or ask your own question.