I have the following TS class:
export class TestClass {
constructor(public options: { a?: number, b?: string, c?: boolean, d?: number }) { }
}
Implementations:
new TestClass({a:123});
new TestClass({b:'hello'});
I am looking for a way to set default values for each of the options properties (a, b, c etc.), so that developer can send only part of the parameters, and the other unassigend parameters will get their defaults.
This option is not available:
constructor(public options: { a?: number = 0, b?: string = '', c?: boolean = false, d?: number = 1 }) { }
because of the "A type literal property cannot have an initializer in TS" restriction.
So the way I implement it now is the following:
constructor(public options: { a?: number, b?: string, c?: boolean, d?: number }) {
!options.a && (options.a = 0),
!options.b && (options.b = ''),
!options.c && (options.c = false),
!options.d && (options.d = 1)
}
My problem is that in the real world the options parameter contains many properties, and I don't want to force each property to be written twice, in the constructor signature and in the constructor's body.
My question is if there is a better way to do it.
Default parameters are used for function/method arguments that are undefined (both for JavaScript and TypeScript), however in your case you have a parameter value with optional properties inside of it. Setting those properties to default values is unfortunately natively not possible.
However, we can slightly improve the syntax of your constructor:
constructor(public options: { a?: number, b?: string, c?: boolean, d?: number }) {
options.a || options.a = 0;
options.b || options.b = '';
options.c || options.c = false;
options.d || options.d = 1;
}
If you are using ES2020+, you can use logical nullish assignment to make your syntax even cleaner:
constructor(public options: { a?: number, b?: string, c?: boolean, d?: number }) {
options.a ??= 0;
options.b ??= '';
options.c ??= false;
options.d ??= 1;
}
More info can be found here.