Is there a way to achieve this using a fresh and sleek typescript assertion function?

Question

Is there a way to achieve this using a fresh and sleek typescript assertion function?

Presented below is the snippet of code:

type Bot = BotActive | BotInactive;

class BotActive {
    public readonly status = "active";

    public interact() {
        console.log("Hello there!");
    }
}

class BotInactive {
    public readonly status = "inactive";
}

function canInteract(bot: Bot) {
    if (bot.status === "inactive") throw Error("Bot cannot interact when it's inactive!");

    bot.interact();
}

Now, I am interested in verifying the status of the bot using a TypeScript assertion function! Here's what I have in mind:

function canInteract(bot: Bot) {
    assertBotStatus(bot, "active");

    bot.interact();
}

Is there a way to achieve this? If so, how would the assertBotStatus function be implemented?

Your help is greatly appreciated!

typescript typescript-types

Answer 1

Answer №1

One way to validate the type of an argument is by using assertion statements -

asserts <<arg>> is <<type>>

.

function assertRobotStateOn(robot: Robot): asserts robot is RobotOn {
  if (robot.state !== 'on') {
    throw new Error('Robot is Off!');
  }
}
function assertRobotStateOff(robot: Robot): asserts robot is RobotOff {
   if (robot.state !== 'off') {
    throw new Error('Robot is On!');
  }
}
function maybeSpeak(robot: Robot) {
    assertRobotStateOn(robot);

    robot.speak();
}
maybeSpeak(new RobotOff());

For more information, you can refer to:

Typescript 3.7 Release Notes
Insightful blog article by Dr. Axel Rauschmayer

Answer 2

One way to validate the type of an argument is by using assertion statements -

asserts <<arg>> is <<type>>

.

function assertRobotStateOn(robot: Robot): asserts robot is RobotOn {
  if (robot.state !== 'on') {
    throw new Error('Robot is Off!');
  }
}
function assertRobotStateOff(robot: Robot): asserts robot is RobotOff {
   if (robot.state !== 'off') {
    throw new Error('Robot is On!');
  }
}
function maybeSpeak(robot: Robot) {
    assertRobotStateOn(robot);

    robot.speak();
}
maybeSpeak(new RobotOff());

For more information, you can refer to:

Typescript 3.7 Release Notes
Insightful blog article by Dr. Axel Rauschmayer

Answer 3

Answer №2

Success! I've discovered a solution using overloaded assertion functions! Check out the code snippet below:

type Android = Active | Inactive;

class Active {
    public readonly status = "active";

    public greet() {
        console.log("Hello there!");
    }
}

class Inactive {
    public readonly status = "inactive"
}

function maybeGreet(android: Android) {
    assertAndroidStatus(android, "active");

    android.greet();
}

function assertAndroidStatus(robot: Android, status: "active"): asserts robot is Active;
function assertAndroidStatus(robot: Android, status: "inactive"): asserts robot is Inactive;
function assertAndroidStatus(robot: Android, state: "active" | "inactive") {
    if (robot.status !== state) throw new Error("Error: Invalid status");
}

Answer 4

Success! I've discovered a solution using overloaded assertion functions! Check out the code snippet below:

type Android = Active | Inactive;

class Active {
    public readonly status = "active";

    public greet() {
        console.log("Hello there!");
    }
}

class Inactive {
    public readonly status = "inactive"
}

function maybeGreet(android: Android) {
    assertAndroidStatus(android, "active");

    android.greet();
}

function assertAndroidStatus(robot: Android, status: "active"): asserts robot is Active;
function assertAndroidStatus(robot: Android, status: "inactive"): asserts robot is Inactive;
function assertAndroidStatus(robot: Android, state: "active" | "inactive") {
    if (robot.status !== state) throw new Error("Error: Invalid status");
}

Is there a way to achieve this using a fresh and sleek typescript assertion function?

Answer №1

Answer №2

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