Is there a way to perform inline type checking in TypeScript without the need for storage?

Question

Is there a way to perform inline type checking in TypeScript without the need for storage?

I am facing a situation where I need to pass an instance of an interface as an argument to a function that accepts any object type, thus lacking type checking. To ensure the object is of the correct type, I usually create an instance and then pass it:

const passMe: ITestInterface = { foo: "bar" };
someFunction(passMe);

However, I am looking for a more concise way to achieve this inline while still enforcing type checking.

// hypothetical syntax example
someFunction({ foo: "bar" } istype ITestInterface);

Is there a straightforward way to do something like the above example in one line?

I've experimented with using 'as', but unfortunately, it doesn't restrict the type. For instance, the following is considered valid:

someFunction({ foo: "bar", hello: true } as ITestInterface);

One workaround could be to modify the someFunction to utilize templating, but this may not always be feasible or practical:

someFunction<TYPE>(arg: TYPE) {
  // function body
}

someFunction<ITestInterface>({foo: "bar"});

typescript typechecking

Answer 1

Answer №1

The particular feature you're searching for, such as "type annotations for arbitrary expressions", is not present in TypeScript at the moment. There is an open proposal regarding this but it is currently marked as "needs proposal". You can show your support by giving it a thumbs up or share your unique ideas if they differ from what's already suggested. However, there doesn't seem to be active development on this feature, so don't expect immediate progress.

There are multiple approaches you can take here, each with its own set of challenges.

As demonstrated, the simplest method is using a type assertion. This helps prevent passing in an entirely different type:

// assertion
someFunction({ foo: "bar" } as ITestInterface); // works fine
someFunction({ unrelatedThing: 1 } as ITestInterface); // results in an error

While this method allows additional properties, it also narrows down types unsafely as shown below:

someFunction({} as ITestInterface); // surprisingly no error due to type narrowing

Another option would be to create a utility function like isType as follows:

// utility function
const isType = <T>(x: T) => x;

This approach offers similar functionality as desired:

someFunction(isType<ITestInterface>({ foo: "bar" })); // works as expected
someFunction(isType<ITestInterface>({ unrelatedThing: 1 })); // generates an error

someFunction(isType<ITestInterface>({ foo: "bar", hello: true })); // returns an error as intended
someFunction(isType<ITestInterface>({})); // results in an error as expected

However, it's worth considering whether this solution aligns with your preferences. Most runtime engines optimize functions like x => x, so performance may not be a concern. It ultimately depends on how you prioritize elegance.

Those are the options available. Hopefully, this information proves helpful. Best of luck!

Link to code

Answer 2

The particular feature you're searching for, such as "type annotations for arbitrary expressions", is not present in TypeScript at the moment. There is an open proposal regarding this but it is currently marked as "needs proposal". You can show your support by giving it a thumbs up or share your unique ideas if they differ from what's already suggested. However, there doesn't seem to be active development on this feature, so don't expect immediate progress.

There are multiple approaches you can take here, each with its own set of challenges.

As demonstrated, the simplest method is using a type assertion. This helps prevent passing in an entirely different type:

// assertion
someFunction({ foo: "bar" } as ITestInterface); // works fine
someFunction({ unrelatedThing: 1 } as ITestInterface); // results in an error

While this method allows additional properties, it also narrows down types unsafely as shown below:

someFunction({} as ITestInterface); // surprisingly no error due to type narrowing

Another option would be to create a utility function like isType as follows:

// utility function
const isType = <T>(x: T) => x;

This approach offers similar functionality as desired:

someFunction(isType<ITestInterface>({ foo: "bar" })); // works as expected
someFunction(isType<ITestInterface>({ unrelatedThing: 1 })); // generates an error

someFunction(isType<ITestInterface>({ foo: "bar", hello: true })); // returns an error as intended
someFunction(isType<ITestInterface>({})); // results in an error as expected

However, it's worth considering whether this solution aligns with your preferences. Most runtime engines optimize functions like x => x, so performance may not be a concern. It ultimately depends on how you prioritize elegance.

Those are the options available. Hopefully, this information proves helpful. Best of luck!

Link to code

Answer 3

Answer №2

To start with, in TypeScript, interfaces are meant to be implemented by a class. It is recommended to use types instead of interfaces or classes for type-checking simple objects. Remember, the 'I' before an interface name signifies it as an interface, so opt for naming conventions like ITest rather than ITestInterface:

// Replace this:
ITestInterface { foo: string }
// With this:
type Test = { foo: string }

Now, diving into the heart of the matter:

If you want to ensure that someFunction always accepts objects of type

Type</code, defining the function like below suffices because TypeScript will catch any instances where it's invoked with another type:</p>

<pre><code>// Example:
const someFunction: (arg: Type) => any = (arg) => { /*...*/ }
// Alternatively
function someFunction(arg: Type): any { /*...*/ }

In cases where you are certain about an argument being of type

Type</code but TypeScript fails to infer it, utilize the <code>as

keyword:

someFunction({foo: 10}); // Error
someFunction({foo: 'bar'});
someFunction({foo: Math.random()<1 ? 'bar' : 10}); // Error
someFunction({foo: Math.random()<1 ? 'bar' : 10} as Type);

Following these steps ensures compile-time type safety in your program. You can experiment with the code above in the TypeScript playground.

If you desire an additional layer of reliability and wish to enforce runtime type safety, you'll need to perform type checks at runtime. While this may lead to slight performance overhead, verifying specific properties on an object can be achieved within a single line in the function definition:

const someSafeFunction(arg: Type): any {
  if (Object.keys(arg).sort().join(',')!='propertyName1,propertyName2') throw new Error('Invalid argument type');
  /* ... */
}

Answer 4

To start with, in TypeScript, interfaces are meant to be implemented by a class. It is recommended to use types instead of interfaces or classes for type-checking simple objects. Remember, the 'I' before an interface name signifies it as an interface, so opt for naming conventions like ITest rather than ITestInterface:

// Replace this:
ITestInterface { foo: string }
// With this:
type Test = { foo: string }

Now, diving into the heart of the matter:

If you want to ensure that someFunction always accepts objects of type

Type</code, defining the function like below suffices because TypeScript will catch any instances where it's invoked with another type:</p>

<pre><code>// Example:
const someFunction: (arg: Type) => any = (arg) => { /*...*/ }
// Alternatively
function someFunction(arg: Type): any { /*...*/ }

In cases where you are certain about an argument being of type

Type</code but TypeScript fails to infer it, utilize the <code>as

keyword:

someFunction({foo: 10}); // Error
someFunction({foo: 'bar'});
someFunction({foo: Math.random()<1 ? 'bar' : 10}); // Error
someFunction({foo: Math.random()<1 ? 'bar' : 10} as Type);

Following these steps ensures compile-time type safety in your program. You can experiment with the code above in the TypeScript playground.

If you desire an additional layer of reliability and wish to enforce runtime type safety, you'll need to perform type checks at runtime. While this may lead to slight performance overhead, verifying specific properties on an object can be achieved within a single line in the function definition:

const someSafeFunction(arg: Type): any {
  if (Object.keys(arg).sort().join(',')!='propertyName1,propertyName2') throw new Error('Invalid argument type');
  /* ... */
}

Is there a way to perform inline type checking in TypeScript without the need for storage?

Answer №1

Answer №2

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