What is the reason behind permitting void functions in the left part of an assignment in Typescript?

Question

What is the reason behind permitting void functions in the left part of an assignment in Typescript?

Take a look at this Typescript snippet:

let action = function (): void {
    //perform actions
};

let result = action();

What makes it suitable for the TypeScript compiler?

typescript

Answer 1

Answer №1

It seems that the reason for this behavior is due to the fact that in Typescript, void is considered a distinct type with specific values like undefined and null, allowing it to be used in assignments.

The potential risk of using values of type void in assignments is concerning as it could result in serious errors. For instance, if a function originally designed to return a meaningful value is changed to return void, there may not be any compile-time error generated.

According to the Typescript 1.4 language specification, the Void type signifies the absence of a value, exclusively accepting null and undefined as its possible values. While Void intersects with Any and encompasses Null and Undefined, its correlation with other types is non-existent.
NOTE: Although declaring variables of type Void may seem unnecessary, their usage becomes inevitable when dealing with generic types or functions.

Answer 2

It seems that the reason for this behavior is due to the fact that in Typescript, void is considered a distinct type with specific values like undefined and null, allowing it to be used in assignments.

The potential risk of using values of type void in assignments is concerning as it could result in serious errors. For instance, if a function originally designed to return a meaningful value is changed to return void, there may not be any compile-time error generated.

According to the Typescript 1.4 language specification, the Void type signifies the absence of a value, exclusively accepting null and undefined as its possible values. While Void intersects with Any and encompasses Null and Undefined, its correlation with other types is non-existent.
NOTE: Although declaring variables of type Void may seem unnecessary, their usage becomes inevitable when dealing with generic types or functions.

Answer 3

Answer №2

What is the reasoning behind allowing void functions in the left side of an assignment in TypeScript?

This allowance is simply because there are no explicit restrictions against it. However, it should be noted that using a void function on the left side doesn't have much practical utility as demonstrated in the example below:

var action = function action(): void {};
var result = action();
result.bar; // Error

Answer 4

What is the reasoning behind allowing void functions in the left side of an assignment in TypeScript?

This allowance is simply because there are no explicit restrictions against it. However, it should be noted that using a void function on the left side doesn't have much practical utility as demonstrated in the example below:

var action = function action(): void {};
var result = action();
result.bar; // Error

Answer 5

Answer №3

Take a look at this code snippet:

// Function to measure performance
function time<T>(f: () => T): T {
  var start = Date.now();
  var result = f();
  console.log('It took ' + (Date.now() - start) + ' to run');
  return f;
}

function doSomething(): void { }

time(doSomething);

If we consider generic functions as specific instances of a template based on type (note: this is not accurate), the variable result will have a type of void when time is called with doSomething. Despite this, there are no catastrophic consequences when dealing with a variable of type void, so it's not imperative to prevent its usage.

Another example:

declare function doNothing(): void;
declare function doOneThing(): void;
declare function doManyThings(): void;

function doSomething(x: number) {
    switch (x) {
        case 0:
            return doNothing();
        case 1:
            return doOneThing();
        default:
            return doManyThings();
    }
}

In this scenario, if we modify the return type of doSomething, we must also update the return types of doNothing, doOneThing, and doManyThings. Once again, we can observe that a void value can be handled without issues. This example is not contrived; it is based on actual TypeScript compiler code.

Let's make some tweaks to the code:

function doSomething(x: number) {
    if (x > 1) {
        return doManyThings();
    } else {
        // result: void!
        var result = x === 0 ? doNothing() : doOneThing();
        return result;
    }
}

Why should such a harmless refactoring trigger an error? What purpose does it serve? If we attempt to access properties of result, it will lead to an error. In cases where we do not need to access any properties, we are likely following correct practices and should not face any issues.

Answer 6

Take a look at this code snippet:

// Function to measure performance
function time<T>(f: () => T): T {
  var start = Date.now();
  var result = f();
  console.log('It took ' + (Date.now() - start) + ' to run');
  return f;
}

function doSomething(): void { }

time(doSomething);

If we consider generic functions as specific instances of a template based on type (note: this is not accurate), the variable result will have a type of void when time is called with doSomething. Despite this, there are no catastrophic consequences when dealing with a variable of type void, so it's not imperative to prevent its usage.

Another example:

declare function doNothing(): void;
declare function doOneThing(): void;
declare function doManyThings(): void;

function doSomething(x: number) {
    switch (x) {
        case 0:
            return doNothing();
        case 1:
            return doOneThing();
        default:
            return doManyThings();
    }
}

In this scenario, if we modify the return type of doSomething, we must also update the return types of doNothing, doOneThing, and doManyThings. Once again, we can observe that a void value can be handled without issues. This example is not contrived; it is based on actual TypeScript compiler code.

Let's make some tweaks to the code:

function doSomething(x: number) {
    if (x > 1) {
        return doManyThings();
    } else {
        // result: void!
        var result = x === 0 ? doNothing() : doOneThing();
        return result;
    }
}

Why should such a harmless refactoring trigger an error? What purpose does it serve? If we attempt to access properties of result, it will lead to an error. In cases where we do not need to access any properties, we are likely following correct practices and should not face any issues.

What is the reason behind permitting void functions in the left part of an assignment in Typescript?

Answer №1

Answer №2

Answer №3

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