In TypeScript, is it possible to indicate that a function will explicitly define a variable?

Question

In TypeScript, is it possible to indicate that a function will explicitly define a variable?

In TypeScript, I am working on creating a class that delays the computation of its information until it is first requested, and then caches it for future use. The basic logic can be summarized as follows.

let foo: string | undefined = undefined;
function defineVariables(): void {
    foo = "foo";
}
function getFoo(): string {
    if (!foo) {
        defineVariables();
    }
    return foo; // type "string | undefined" not assignable to type "string"
}

However, there is an error in this code because the compiler cannot determine that after defineVariables() is called, foo will no longer be undefined. Is there a way to explicitly declare this to the compiler, or do I need to consider a different approach? Searching for solutions to this issue online yields results that are not quite relevant due to the general terms used.

It is worth noting that while this example may seem unnecessary, imagine a scenario where defineVariables() involves complex calculations and defines multiple variables, with getFoo() being called multiple times.

Edit: Following @ksav's suggestion, using return foo as string; suppresses the error, but I am unsure if it fully addresses the underlying problem. I am interested in exploring if there is a more appropriate solution within the defineVariables() function.

typescript

Answer 1

Answer №1

One issue you're facing is relying on side effects instead of function return types in your code. Typescript struggles with side effects and they can lead to bugs.

There are several ways you could address this:

You have the option to simply use return foo as string to explicitly state that foo will be a string at that point. While this is straightforward, it essentially overrides Typescript and could potentially break your code in the future if the codepath where foo is set changes.
Another approach is to assert at runtime that foo has a defined value, or throw an exception if it doesn't:

let foo: string | undefined = undefined;
function defineVariables(): void {
    foo = "foo";
}
function getFoo(): string {
    if (!foo) {
        defineVariables();
    }
    if (foo) {
        return foo;
    }
    throw new Error("Foo failed to define");
}

While effective, this method adds more runtime overhead and won't catch errors during compile time if the defineVariables() function no longer guarantees foo's definition.

Alternatively:

My personal suggestion would be to refactor your code to focus on return values rather than side effects:

let foo: string | undefined = undefined;
function defineVariables(): string {
    foo = "foo";
    // additional logic here
    return foo;
}

function getFoo(): string {
    if (!foo) {
        return defineVariables();
    }
    return foo;
}

Although this example may seem simplistic with just one variable, it limits the usage of foo in getFoo to a context where it will always be inferred as not undefined.

Answer 2

One issue you're facing is relying on side effects instead of function return types in your code. Typescript struggles with side effects and they can lead to bugs.

There are several ways you could address this:

You have the option to simply use return foo as string to explicitly state that foo will be a string at that point. While this is straightforward, it essentially overrides Typescript and could potentially break your code in the future if the codepath where foo is set changes.
Another approach is to assert at runtime that foo has a defined value, or throw an exception if it doesn't:

let foo: string | undefined = undefined;
function defineVariables(): void {
    foo = "foo";
}
function getFoo(): string {
    if (!foo) {
        defineVariables();
    }
    if (foo) {
        return foo;
    }
    throw new Error("Foo failed to define");
}

While effective, this method adds more runtime overhead and won't catch errors during compile time if the defineVariables() function no longer guarantees foo's definition.

Alternatively:

My personal suggestion would be to refactor your code to focus on return values rather than side effects:

let foo: string | undefined = undefined;
function defineVariables(): string {
    foo = "foo";
    // additional logic here
    return foo;
}

function getFoo(): string {
    if (!foo) {
        return defineVariables();
    }
    return foo;
}

Although this example may seem simplistic with just one variable, it limits the usage of foo in getFoo to a context where it will always be inferred as not undefined.

Answer 3

Answer №2

To accomplish this, you can utilize an asserts return type. It's worth noting that I shifted the if statement from the getter to the initializer because Typescript may not properly handle control-flow narrowing otherwise.

class Foo {
    private foo: string | undefined = undefined;

    public getFoo(): string {
        this.ensureFoo();
        return this.foo;
    }

    private ensureFoo(): asserts this is Record<'foo', string> {
        if(!this.foo) {
            this.foo = 'bar';
        }
    }
}

Playground Link

Answer 4

To accomplish this, you can utilize an asserts return type. It's worth noting that I shifted the if statement from the getter to the initializer because Typescript may not properly handle control-flow narrowing otherwise.

class Foo {
    private foo: string | undefined = undefined;

    public getFoo(): string {
        this.ensureFoo();
        return this.foo;
    }

    private ensureFoo(): asserts this is Record<'foo', string> {
        if(!this.foo) {
            this.foo = 'bar';
        }
    }
}

Playground Link

In TypeScript, is it possible to indicate that a function will explicitly define a variable?

Answer №1

Answer №2

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