Property '…' has no initializer and is not definitely assigned in the constructor

Question

in my Angular app i have a component:

import { MakeService } from './../../services/make.service';
import { Component, OnInit } from '@angular/core';

@Component({
  selector: 'app-vehicle-form',
  templateUrl: './vehicle-form.component.html',
  styleUrls: ['./vehicle-form.component.css']
})
export class VehicleFormComponent implements OnInit {
  makes: any[];
  vehicle = {};

  constructor(private makeService: MakeService) { }

  ngOnInit() {
    this.makeService.getMakes().subscribe(makes => { this.makes = makes
      console.log("MAKES", this.makes);
    });
  }

  onMakeChange(){
    console.log("VEHICLE", this.vehicle);
  }
}

but in the "makes" property I have a mistake. I dont know what to do with it...

Answer 1

I think you are using the latest version of TypeScript. Please see the section "Strict Class Initialization" in the link.

There are two ways to fix this:

A. If you are using VSCode you need to change the TS version that the editor use.

B. Just initialize the array when you declare it inside the constructor,

makes: any[] = [];

constructor(private makeService: MakeService) { 
   // Initialization inside the constructor
   this.makes = [];
}

Answer 2

Just go to tsconfig.json and set

"strictPropertyInitialization": false

to get rid of the compilation error.

Otherwise you need to initialize all your variables which is a little bit annoying

Answer 3

It is because TypeScript 2.7 includes a strict class checking where all the properties should be initialized in the constructor. A workaround is to add the ! as a postfix to the variable name:

makes!: any[];

Answer 4

We may get the message Property has no initializer and is not definitely assigned in the constructor when adding some configuration in the tsconfig.json file so as to have an Angular project compiled in strict mode:

"compilerOptions": {
  "strict": true,
  "noImplicitAny": true,
  "noImplicitThis": true,
  "alwaysStrict": true,
  "strictNullChecks": true,
  "strictFunctionTypes": true,
  "strictPropertyInitialization": true,

Indeed the compiler then complains that a member variable is not defined before being used.

For an example of a member variable that is not defined at compile time, a member variable having an @Input directive:

@Input() userId: string;

We could silence the compiler by stating the variable may be optional:

@Input() userId?: string;

But then, we would have to deal with the case of the variable not being defined, and clutter the source code with some such statements:

if (this.userId) {
} else {
}

Instead, knowing the value of this member variable would be defined in time, that is, it would be defined before being used, we can tell the compiler not to worry about it not being defined.

The way to tell this to the compiler is to add the ! definite assignment assertion operator, as in:

@Input() userId!: string;

Now, the compiler understands that this variable, although not defined at compile time, shall be defined at run-time, and in time, before it is being used.

It is now up to the application to ensure this variable is defined before being used.

As an an added protection, we can assert the variable is being defined, before we use it.

We can assert the variable is defined, that is, the required input binding was actually provided by the calling context:

private assertInputsProvided(): void {
  if (!this.userId) {
    throw (new Error("The required input [userId] was not provided"));
  }
}

public ngOnInit(): void {
  // Ensure the input bindings are actually provided at run-time
  this.assertInputsProvided();
}

Knowing the variable was defined, the variable can now be used:

ngOnChanges() {
  this.userService.get(this.userId)
    .subscribe(user => {
      this.update(user.confirmedEmail);
    });
}

Note that the ngOnInit method is called after the input bindings attempt, this, even if no actual input was provided to the bindings.

Whereas the ngOnChanges method is called after the input bindings attempt, and only if there was actual input provided to the bindings.

Answer 5

Go to your tsconfig.json file and change "noImplicitReturns": false then add "strictPropertyInitialization": false to your tsconfig.json file under "compilerOptions" property. Here is what my tsconfig.json file looks like

tsconfig.json

{
  ...
  "compilerOptions": {
        ....
        "noImplicitReturns": false,
        ....
        "strictPropertyInitialization": false
  },
  "angularCompilerOptions": {
     ......
  }  
}

Hope this will help !! Good Luck

Answer 6

The error is legitimate and may prevent your app from crashing. You typed makes as an array but it can also be undefined.

You have 2 options (instead of disabling the typescript's reason for existing...):

1. In your case the best is to type makes as possibily undefined.

makes?: any[]
// or
makes: any[] | undefined

So the compiler will inform you whenever you try to access to makes that it could be undefined. Otherwise, if the // <-- Not ok lines below were executed before getMakes finished or if getMakes failed, your app would crash and a runtime error would be thrown. That's definitely not what you want.

makes[0] // <-- Not ok
makes.map(...) // <-- Not ok

if (makes) makes[0] // <-- Ok
makes?.[0] // <-- Ok
(makes ?? []).map(...) // <-- Ok

2. You can assume that it will never fail and that you will never try to access it before initialization by writing the code below (risky!). So the compiler won't take care about it.

makes!: any[] 

Answer 7

Update for 2021 :

there is property like "strictPropertyInitialization"

Just go to tsconfig.json and set

"strict": false

to get rid of the compilation error.

Otherwise you need to initialize all your variables which is a little bit annoying.

reason behind this error is :

• typescript is a kind of more Secure lang as compare to javascript.
• although this security is enhance by enabling strict feature .So every time when you initialize a variable typescript wants them to assign a value.

Answer 8

You either need to disable the --strictPropertyInitialization that Sajeetharan referred to, or do something like this to satisfy the initialization requirement:

makes: any[] = [];

Answer 9

You can also do the following if you really don't want to initialise it.

makes?: any[];

Answer 10

As of TypeScript 2.7.2, you are required to initialise a property in the constructor if it was not assigned to at the point of declaration.

If you are coming from Vue, you can try the following:

• Add "strictPropertyInitialization": true to your tsconfig.json

• If you are unhappy with disabling it you could also try this makes: any[] | undefined. Doing this requires that you access the properties with null check (?.) operator i.e. this.makes?.length

• You could as well try makes!: any[];, this tells TS that the value will be assigned at runtime.

Answer 11

Get this error at the time of adding Node in my Angular project -

TSError: ? Unable to compile TypeScript: (path)/base.api.ts:19:13 - error TS2564: Property 'apiRoot Path' has no initializer and is not definitely assigned in the constructor.

private apiRootPath: string;

Solution -

Added "strictPropertyInitialization": false in 'compilerOptions' of tsconfig.json.

my package.json -

"dependencies": {
    ...
    "@angular/common": "~10.1.3",
    "@types/express": "^4.17.9",
    "express": "^4.17.1",
    ...
}

Ref URL - https://www.ryadel.com/en/ts2564-ts-property-has-no-initializer-typescript-error-fix-visual-studio-2017-vs2017/

Answer 12

If you want to initialize an object based on an interface you can initialize it empty with following statement.

myObj: IMyObject = {} as IMyObject;

Answer 13

When you upgrade using [email protected] , its compiler strict the rules follows for array type declare inside the component class constructor.

For fix this issue either change the code where are declared in the code or avoid to compiler to add property "strictPropertyInitialization": false in the "tsconfig.json" file and run again npm start .

Angular web and mobile Application Development you can go to www.jtechweb.in

Answer 14

Can't you just use a Definite Assignment Assertion? (See https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-7.html#definite-assignment-assertions)

i.e. declaring the property as makes!: any[]; The ! assures typescript that there definitely will be a value at runtime.

Sorry I haven't tried this in angular but it worked great for me when I was having the exact same problem in React.

Answer 15

Another way to fix in the case when the variable must remain uninitialized (and it is dealt with at the run time) is to add undefined to the type (this is actually suggested by VC Code). Example:

@Input() availableData: HierarchyItem[] | undefined;
@Input() checkableSettings: CheckableSettings | undefined;

Depending on actual usage, this might lead to other issues, so I think the best approach is to initialize the properties whenever possible.

Answer 16

change the

fieldname?: any[]; 

to this:

fieldname?: any; 

Answer 17

You can declare property in constructor like this:

export class Test {
constructor(myText:string) {
this.myText= myText;
} 

myText: string ;
}

Answer 18

Next to variables "?" You can fix it by putting it.

Example:

--------->id?:number --------->name?:string

Answer 19

I was using generic types and faced the same issue along with the issue that I could not initialize the property as it was generic it could be any type

class HoldAnything<TypeOfData>{
  data:TypeOfData;
 
}

const holdNumber = new HoldAnything<number>();
holdNumber.data =123;

const holdString = new HoldAnything<string>()
holdString.data="arvind";

So for this I put strictPropertyInitialization to false in tsconfig.json file.

Got clarification from here Strict Class Initialization