TypeScript strict class / interface

Question

I use TypeScript version 2.3.4. I want to write a function that accepts an object that must have specified fields. But this object should not contain any other fields. How can I achieve this?

Now this works only if I define object inline. But if I use another object with extra fields - the compiler will allow it. What is totally wrong.

Example:

function foo(arg: { a: string }) { // there is tons of fields actually
  // ...
}

foo ({a: "qwerty"}); // No Error - ok

foo ({a: "qwerty", b: 123}); // Error - ok

let bar = {
  a: "qwerty",
  b: 123
};

foo (bar); // No Error - NOT OK !!!

The same code can be writed with interfaces, classes, type declarations - it's the same problem.

Now I have to extract the fields from the object manually to make sure that there are no extra fields. But I can't spread this solution on ~1000 functions (I really need this) all over the code - it's too messy. I creating API wrapper and I need to ensure that there are no additional or wrong fields passed to the server.

Looks like it is intentional. See this issue and this for additional discussion. — Saravana
– Saravana, Commented Jun 20, 2017 at 7:09
Saravana, yeah, I read this, but there is no answer for my question in theese discussions. I think is will be better to avoid strictness for interfaces, but leave it for classes. — Harry Burns
– Harry Burns, Commented Jun 20, 2017 at 7:29

Stefan Steiger · Accepted Answer · 2017-06-23 15:17:09Z

6

The feature you are asking for is known as "exact types".
It's being considered, that is, neither rejected nor accepted, and the discussion still goes on.

edited Jun 23, 2017 at 15:17

Stefan Steiger

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answered Jun 20, 2017 at 7:08

artem

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Comments

user6038265 · Accepted Answer · 2017-06-23 15:15:25Z

3

Necromancing.
It's coming with typescript 2.4:

type foo = {
  a:string;
  b:number;
  opt?:string;
}


function test(obj:foo)
{}

test({ a:"", b:123, e:"produceError"});

And to enforce it's an object, if all parameters are optional:

function test(obj:foo & object)
{}

And if you want to pass either a string or a another object-type:

function test(obj: string | foo & object)
{}

answered Jun 23, 2017 at 15:15

user6038265

Comments

smnbbrv · Accepted Answer · 2017-06-20 06:46:07Z

2

This is how it works by design and I'm not sure you can work this around. See the docs for more info.

What you also do wrong - you cannot be 100% sure about what's sent to the server. As long as browser does not know anything about TS, some library can inject into any request whatever it needs with e.g. by overwriting XmlHttpRequest methods (this is what e.g. at least angular's zone.js does).

Another way to easily break your intentions is as simple as using <any> in front of any parameter you pass.

TypeScript is intended to improve your development process but I don't think it can be used for the needs like yours. This is usually covered by writing the proper tests.

answered Jun 20, 2017 at 6:46

smnbbrv

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1 Comment

Harry Burns Over a year ago

Of course I'm know that TS compiled to JS where no classes or type checks! I need to secure my method from other developers in development process. This all about development. Real case example: codepen.io/HarryBurns/pen/LLWoVa.typescript

Rahul Singh · Accepted Answer · 2017-06-20 06:48:43Z

2

There "is" a way, but you have to implement it yourself. It's called a "User Defined Type Guard" and it looks like this:

interface Test {
    prop: number;
}

function isTest(arg: any): arg is Test {
    return arg && arg.prop && typeof(arg.prop) == 'number';
}

answered Jun 20, 2017 at 6:48

Rahul Singh

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1 Comment

Harry Burns Over a year ago

It's better to do with Reflection (see blog.wolksoftware.com/…), so it will be shorter, something like check<MyType>(myObject). Or, in my case it's pick<MyType>(myObject). Works like underscore/lodash pick function.

artur grzesiak · Accepted Answer · 2021-06-13 14:41:08Z

1

With conditional types it is possible to check if two types extend each other - kind of equality check for types. Below a possible solution:

type IsEqual<A,B> = A extends B ? B extends A ? true : false : false
type FooParams = { a: string }

function foo<A extends FooParams>(arg: IsEqual<A, FooParams> extends true ? A : never) {
  // ...
}

foo ({a: "qwerty"}); // No Error - ok

foo ({a: "qwerty", b: 123}); // Error - ok

let bar = {
  a: "qwerty",
  b: 123
};

foo (bar); // Error - ok

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answered Jun 13, 2021 at 14:41

artur grzesiak

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