CODEWITHSUNDEEP
typescript
grace
grace

Reputation: 270

Wants to force a function of a specific type, but want to have a default value in TypeScript

I want to force a function of a specific type, but want to have a default value for some functions.

And I tried with simple code like below

type MyFunc = (str: string) => number;

const myAFunc: MyFunc = (str) => Number(str);
const myBFunc: MyFunc = (str = '9999') => Number(str);

myAFunc('a');
myAFunc(); // wants to occur error
myBFunc('b');
myBFunc(); // wants to NOT occur error

https://www.typescriptlang.org/play?#code/C4TwDgpgBAsiBiBXAdgYygXigCgM7ACcAuKfAgS2QHMBKTAPimUQFsAjCAgbgCgfUA9snxQWIAIJI0JOFPRY8hOhkYA5VhwKKCNXoOHBRIAEJyZCFPJxlMUAOQBOJw7vK1Gztt18xky9jsAQ1deXzlsXSgAeiioAHdA5GBcKGABKAFUVEQCKE4CAQIeMVN-OzYQ4pNwyJj4xOTU9NUAeQAVDKycvIICgiA

But it's not working.

Is there some way to solve it?

Upvotes: 3

Views: 361

Answers (2)

Hyunjune Kim
Hyunjune Kim

Reputation: 485

If you want to make it optional to pass anything to the function, add a ? for parameter in the type.

type MyFunc = (str?: string) => number;

const myAFunc: MyFunc = (str) => Number(str);
const myBFunc: MyFunc = (str = '9999') => Number(str);

myAFunc(); // NaN
myBFunc(); // 9999

Upvotes: 0

gaborp
gaborp

Reputation: 714

A and B func has different types, don't force them into same type:

type MyAFunc = (str: string) => number;
type MyBFunc = (str?: string) => number;

const myAFunc: MyAFunc = (str) => Number(str);
const myBFunc: MyBFunc = (str = '9999') => Number(str);

myAFunc('a');
myAFunc(); // wants to occur error
myBFunc('b');
myBFunc(); // wants to NOT occur error

Upvotes: 3

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