Error with declaring TypeScript class due to private variable

Question

Error with declaring TypeScript class due to private variable

When defining a TypeScript class like this:

export class myClass {
  constructor(public aVariable: number) {}
  private aPrivateVariable: number;
}

and trying to initialize it with the following code:

let someVar: myClass[] = [{
  aVariable: 3
}, {
  aVariable: 2
}];

An error, typically in VS Code, is thrown stating:

The property 'aPrivateVariable' is missing in type '{ aVariable: number; }'.

This brings up the question: Why am I unable to do this?

Thank you.

javascript typescript visual-studio-code

Answer 1

Answer №1

Picture a scenario where your class contains a method that utilizes the given field:

 export class myCustomClass {
  constructor(public someValue: number) {}
  private anotherValue: number;
  performTest(){ }
 }

If you were to execute the following:

 const exampleInstance: myCustomClass = {
    someValue: 1,
    performTest() {
      alert.anotherValue + 1);
    }
 };

 exampleInstance.performTest();

This would not work as expected. Therefore, it is necessary to include private properties as well.

 let arrayOfObjects: myCustomClass[] = [{
  someValue: 3,
  anotherValue: 2,
 }, {
  someValue: 2,
  anotherValue: 3
 }];

Answer 2

Picture a scenario where your class contains a method that utilizes the given field:

 export class myCustomClass {
  constructor(public someValue: number) {}
  private anotherValue: number;
  performTest(){ }
 }

If you were to execute the following:

 const exampleInstance: myCustomClass = {
    someValue: 1,
    performTest() {
      alert.anotherValue + 1);
    }
 };

 exampleInstance.performTest();

This would not work as expected. Therefore, it is necessary to include private properties as well.

 let arrayOfObjects: myCustomClass[] = [{
  someValue: 3,
  anotherValue: 2,
 }, {
  someValue: 2,
  anotherValue: 3
 }];

Answer 3

Answer №2

When defining objects like { aVariable: 2}, they are not instantiated as instances of the myClass. The issue arises because aPrivateVariable is not being initialized since the constructor of myClass is never invoked, and the object structure does not align with what myClass expects.

As mentioned in the typescript handbook's section on type compatibility:

Private and protected members within a class impact their compatibility. When checking compatibility for an instance of a class, if the target type includes a private member, the source type must also have a private member from the same class. Similarly, this rule applies for instances with protected members. This ensures that a class can be compatible with its superclass, but not with classes from different inheritance hierarchies even if they share similar structures.

One way to address this issue is by instantiating each object using functions like map()


const someVar: myClass[] = [2,3,1].map(num => myClass(num));

Alternatively, if you are certain about the object's structure, you can utilize type coercion:


let someVar: myClass[] = [] as myClass[]

It's important to exercise caution with this approach as it doesn't ensure that the array elements match the model of the class.

Answer 4

When defining objects like { aVariable: 2}, they are not instantiated as instances of the myClass. The issue arises because aPrivateVariable is not being initialized since the constructor of myClass is never invoked, and the object structure does not align with what myClass expects.

As mentioned in the typescript handbook's section on type compatibility:

Private and protected members within a class impact their compatibility. When checking compatibility for an instance of a class, if the target type includes a private member, the source type must also have a private member from the same class. Similarly, this rule applies for instances with protected members. This ensures that a class can be compatible with its superclass, but not with classes from different inheritance hierarchies even if they share similar structures.

One way to address this issue is by instantiating each object using functions like map()


const someVar: myClass[] = [2,3,1].map(num => myClass(num));

Alternatively, if you are certain about the object's structure, you can utilize type coercion:


let someVar: myClass[] = [] as myClass[]

It's important to exercise caution with this approach as it doesn't ensure that the array elements match the model of the class.

Answer 5

Answer №3

class MyClass {
   constructor(public myVar: number) {}
   private myPrivateVar: number;
}

The proper way to initialize your class is as follows:

const someArray: MyClass[] = [
    new MyClass(3),
    new MyClass(2),
]

If you fail to initialize your class and attempt to access any methods from MyClass, you will receive an object cast as MyClass.

Answer 6

class MyClass {
   constructor(public myVar: number) {}
   private myPrivateVar: number;
}

The proper way to initialize your class is as follows:

const someArray: MyClass[] = [
    new MyClass(3),
    new MyClass(2),
]

If you fail to initialize your class and attempt to access any methods from MyClass, you will receive an object cast as MyClass.

Answer 7

Answer №4

When you encounter the error message stating that { aVariable: number; } isn't compatible with myClass,

According to the documentation,

Private and protected members within a class impact their compatibility. In order for an instance of a class to be considered compatible, the target type must contain a private member if the source type contains one from the same class. The same rule applies for instances with protected members. This design allows a class to be assignment compatible with its super class, but not with classes from a different inheritance hierarchy even if they have a similar structure.

If your intention is to use myClass as an interface for assigning plain objects that match the structure of the myClass class, you should create a separate interface:

interface IMyClass {
  aVariable: number;
}

export class myClass implements IMyClass {...}

let someVar: IMyClass[] = [{
  aVariable: 3
}];

After creating the interface, proceed to initialize it.

It's important to note that the above code snippet does not actually instantiate the class. If the intention is to populate someVar with instances of myClass, the type error indicates an issue. Instances should be created explicitly using new:

let someVar: myClass[] = [new myClass(3)];

Answer 8

When you encounter the error message stating that { aVariable: number; } isn't compatible with myClass,

According to the documentation,

Private and protected members within a class impact their compatibility. In order for an instance of a class to be considered compatible, the target type must contain a private member if the source type contains one from the same class. The same rule applies for instances with protected members. This design allows a class to be assignment compatible with its super class, but not with classes from a different inheritance hierarchy even if they have a similar structure.

If your intention is to use myClass as an interface for assigning plain objects that match the structure of the myClass class, you should create a separate interface:

interface IMyClass {
  aVariable: number;
}

export class myClass implements IMyClass {...}

let someVar: IMyClass[] = [{
  aVariable: 3
}];

After creating the interface, proceed to initialize it.

It's important to note that the above code snippet does not actually instantiate the class. If the intention is to populate someVar with instances of myClass, the type error indicates an issue. Instances should be created explicitly using new:

let someVar: myClass[] = [new myClass(3)];

Answer 9

Answer №5

Remember to include the class declaration when initializing an array of objects:

const items: MyClass[] = [new MyClass(3), new MyClass(2)];

Answer 10

Remember to include the class declaration when initializing an array of objects:

const items: MyClass[] = [new MyClass(3), new MyClass(2)];

Answer 11

Answer №6

The proper method for initializing this class is as follows:

export class ExampleClass {
  constructor(public aNumber: number) {}
  private aPrivateNumber: number;
}

let someVariable: ExampleClass = new ExampleClass(3);

Answer 12

The proper method for initializing this class is as follows:

export class ExampleClass {
  constructor(public aNumber: number) {}
  private aPrivateNumber: number;
}

let someVariable: ExampleClass = new ExampleClass(3);

Error with declaring TypeScript class due to private variable

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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