How to use 'unknown' without instanceof in a meaningful way?

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Since TypeScript 3.0 the new top type unknown is available as a type-safe counterpart of any.

I really love this type, as it makes the code even more robust.

My main problem with this is, that TypeScript (obviously) does not provide an instanceof-feature at runtime for interfaces. If it would (be possible), this code would work quite well:

class Test {



  doSomething(value: unknown) {

    if(value instanceof MySuperInterface) {

      return value.superFunction();

    }



    return false;

  }



}

My question

How do I use this type in a meaningful way, without the need of checking each property of my interface?

asked Nov 23 '18 at 14:10

scipper

1,4631424

1

In conjunction with discriminated unions maybe?

– Fenton
Nov 23 '18 at 14:17

Pick a couple of properties that are significant and use custom type guard to narrow unknown to your specific interface type ?

– Titian Cernicova-Dragomir
Nov 23 '18 at 14:19

add a comment |

Since TypeScript 3.0 the new top type unknown is available as a type-safe counterpart of any.

I really love this type, as it makes the code even more robust.

My main problem with this is, that TypeScript (obviously) does not provide an instanceof-feature at runtime for interfaces. If it would (be possible), this code would work quite well:

class Test {



  doSomething(value: unknown) {

    if(value instanceof MySuperInterface) {

      return value.superFunction();

    }



    return false;

  }



}

My question

How do I use this type in a meaningful way, without the need of checking each property of my interface?

asked Nov 23 '18 at 14:10

scipper

1,4631424

1

In conjunction with discriminated unions maybe?

– Fenton
Nov 23 '18 at 14:17

Pick a couple of properties that are significant and use custom type guard to narrow unknown to your specific interface type ?

– Titian Cernicova-Dragomir
Nov 23 '18 at 14:19

add a comment |

Since TypeScript 3.0 the new top type unknown is available as a type-safe counterpart of any.

I really love this type, as it makes the code even more robust.

My main problem with this is, that TypeScript (obviously) does not provide an instanceof-feature at runtime for interfaces. If it would (be possible), this code would work quite well:

class Test {



  doSomething(value: unknown) {

    if(value instanceof MySuperInterface) {

      return value.superFunction();

    }



    return false;

  }



}

My question

How do I use this type in a meaningful way, without the need of checking each property of my interface?

asked Nov 23 '18 at 14:10

scipper

1,4631424

Since TypeScript 3.0 the new top type unknown is available as a type-safe counterpart of any.

I really love this type, as it makes the code even more robust.

My main problem with this is, that TypeScript (obviously) does not provide an instanceof-feature at runtime for interfaces. If it would (be possible), this code would work quite well:

class Test {



  doSomething(value: unknown) {

    if(value instanceof MySuperInterface) {

      return value.superFunction();

    }



    return false;

  }



}

My question

How do I use this type in a meaningful way, without the need of checking each property of my interface?

typescript

asked Nov 23 '18 at 14:10

scipper

1,4631424

asked Nov 23 '18 at 14:10

scipper

1,4631424

asked Nov 23 '18 at 14:10

scipper

1,4631424

asked Nov 23 '18 at 14:10

scipper

1,4631424

asked Nov 23 '18 at 14:10

scipper

1,4631424

1

In conjunction with discriminated unions maybe?

– Fenton
Nov 23 '18 at 14:17

Pick a couple of properties that are significant and use custom type guard to narrow unknown to your specific interface type ?

– Titian Cernicova-Dragomir
Nov 23 '18 at 14:19

add a comment |

1

In conjunction with discriminated unions maybe?

– Fenton
Nov 23 '18 at 14:17

Pick a couple of properties that are significant and use custom type guard to narrow unknown to your specific interface type ?

– Titian Cernicova-Dragomir
Nov 23 '18 at 14:19

In conjunction with discriminated unions maybe?

– Fenton
Nov 23 '18 at 14:17

Pick a couple of properties that are significant and use custom type guard to narrow unknown to your specific interface type ?

– Titian Cernicova-Dragomir
Nov 23 '18 at 14:19

add a comment |

1 Answer
1

active

oldest

votes

Javascipt does not check for interface type at runtime.

Typescript: instanceof check on interface

But, you can use unknown type with classes as shown below:

interface MySuperInterface {

    superfunction(): string;

}



class MySuperImplementation implements MySuperInterface {

    superfunction(): string {

        return "Hello from super function";

    }

}



class Greeter {

    greeting: string;



    constructor(message: string) {

        this.greeting = message;

    }



    greet() {

        return "Hello, " + this.greeting;

    }



    testSuperFunction(value: unknown) {

        if (value instanceof MySuperImplementation) {

            return value.superfunction();

        }

        else {

            return "Error";

        }

    }

}



let greeter = new Greeter("world");



const sfunc = new MySuperImplementation();



console.log(greeter.testSuperFunction(sfunc));

https://www.typescriptlang.org/play/index.html#src=interface%20MySuperInterface%20%7B%0D%0A%20%20%20%20superfunction()%3A%20string%3B%0D%0A%7D%0D%0A%0D%0Aclass%20MySuperImplementation%20implements%20MySuperInterface%20%7B%0D%0A%20%20%20%20superfunction()%3A%20string%20%7B%0D%0A%20%20%20%20%20%20%20%20return%20%22Hello%20from%20super%20function%22%3B%0D%0A%20%20%20%20%7D%0D%0A%7D%0D%0A%0D%0Aclass%20Greeter%20%7B%0D%0A%20%20%20%20greeting%3A%20string%3B%0D%0A%0D%0A%20%20%20%20constructor(message%3A%20string)%20%7B%0D%0A%20%20%20%20%20%20%20%20this.greeting%20%3D%20message%3B%0D%0A%20%20%20%20%7D%0D%0A%0D%0A%20%20%20%20greet()%20%7B%0D%0A%20%20%20%20%20%20%20%20return%20%22Hello%2C%20%22%20%2B%20this.greeting%3B%0D%0A%20%20%20%20%7D%0D%0A%0D%0A%20%20%20%20testSuperFunction(value%3A%20unknown)%20%7B%0D%0A%20%20%20%20%20%20%20%20if%20(value%20instanceof%20MySuperImplementation)%20%7B%0D%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20value.superfunction()%3B%0D%0A%20%20%20%20%20%20%20%20%7D%0D%0A%20%20%20%20%20%20%20%20else%20%7B%0D%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20%22Error%22%3B%0D%0A%20%20%20%20%20%20%20%20%7D%0D%0A%20%20%20%20%7D%0D%0A%7D%0D%0A%0D%0Alet%20greeter%20%3D%20new%20Greeter(%22world%22)%3B%0D%0A%0D%0Aconst%20sfunc%20%3D%20new%20MySuperImplementation()%3B%0D%0A%0D%0Aconsole.log(greeter.testSuperFunction(sfunc))%3B

answered Nov 23 '18 at 17:31

SSK

62212

add a comment |

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1 Answer
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active

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1 Answer
1

active

oldest

votes

Javascipt does not check for interface type at runtime.

Typescript: instanceof check on interface

But, you can use unknown type with classes as shown below:

interface MySuperInterface {

    superfunction(): string;

}



class MySuperImplementation implements MySuperInterface {

    superfunction(): string {

        return "Hello from super function";

    }

}



class Greeter {

    greeting: string;



    constructor(message: string) {

        this.greeting = message;

    }



    greet() {

        return "Hello, " + this.greeting;

    }



    testSuperFunction(value: unknown) {

        if (value instanceof MySuperImplementation) {

            return value.superfunction();

        }

        else {

            return "Error";

        }

    }

}



let greeter = new Greeter("world");



const sfunc = new MySuperImplementation();



console.log(greeter.testSuperFunction(sfunc));

answered Nov 23 '18 at 17:31

SSK

62212

add a comment |

Javascipt does not check for interface type at runtime.

Typescript: instanceof check on interface

But, you can use unknown type with classes as shown below:

interface MySuperInterface {

    superfunction(): string;

}



class MySuperImplementation implements MySuperInterface {

    superfunction(): string {

        return "Hello from super function";

    }

}



class Greeter {

    greeting: string;



    constructor(message: string) {

        this.greeting = message;

    }



    greet() {

        return "Hello, " + this.greeting;

    }



    testSuperFunction(value: unknown) {

        if (value instanceof MySuperImplementation) {

            return value.superfunction();

        }

        else {

            return "Error";

        }

    }

}



let greeter = new Greeter("world");



const sfunc = new MySuperImplementation();



console.log(greeter.testSuperFunction(sfunc));

answered Nov 23 '18 at 17:31

SSK

62212

add a comment |

Javascipt does not check for interface type at runtime.

Typescript: instanceof check on interface

But, you can use unknown type with classes as shown below:

interface MySuperInterface {

    superfunction(): string;

}



class MySuperImplementation implements MySuperInterface {

    superfunction(): string {

        return "Hello from super function";

    }

}



class Greeter {

    greeting: string;



    constructor(message: string) {

        this.greeting = message;

    }



    greet() {

        return "Hello, " + this.greeting;

    }



    testSuperFunction(value: unknown) {

        if (value instanceof MySuperImplementation) {

            return value.superfunction();

        }

        else {

            return "Error";

        }

    }

}



let greeter = new Greeter("world");



const sfunc = new MySuperImplementation();



console.log(greeter.testSuperFunction(sfunc));

answered Nov 23 '18 at 17:31

SSK

62212

Javascipt does not check for interface type at runtime.

Typescript: instanceof check on interface

But, you can use unknown type with classes as shown below:

interface MySuperInterface {

    superfunction(): string;

}



class MySuperImplementation implements MySuperInterface {

    superfunction(): string {

        return "Hello from super function";

    }

}



class Greeter {

    greeting: string;



    constructor(message: string) {

        this.greeting = message;

    }



    greet() {

        return "Hello, " + this.greeting;

    }



    testSuperFunction(value: unknown) {

        if (value instanceof MySuperImplementation) {

            return value.superfunction();

        }

        else {

            return "Error";

        }

    }

}



let greeter = new Greeter("world");



const sfunc = new MySuperImplementation();



console.log(greeter.testSuperFunction(sfunc));

answered Nov 23 '18 at 17:31

SSK

62212

answered Nov 23 '18 at 17:31

SSK

62212

answered Nov 23 '18 at 17:31

SSK

62212

answered Nov 23 '18 at 17:31

SSK

62212

add a comment |

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