I have code similar to the following:

export type Bar = {
    baz: number
};

export type FooConstructorArguments = {
    config: Bar
};

class Foo {
    construction({
        config: {
            baz = 1
        }
    }: FooConstructorArguments) {
        let c = config;
        let b = baz;
    }
}

But neither config nor baz are accessible. What is the proper way to utilize nested typed argument defaults within TypeScript?

CodePudding user response：

baz is accessible. config isn't because it's just part of the destructuring path to baz, not a target of the destructuring pattern. If you want both config and baz, you'll need to just pick out config in the parameter list, then pick baz from it after:

class Foo {
    construction({ config }: FooConstructorArguments) {
        const { baz = 1} = config;
        let c = config;
        let b = baz;
    }
}

Playground link

or if you don't really need baz and just need b:

class Foo {
    construction({ config }: FooConstructorArguments) {
        let c = config;
        let b = config.baz ?? 1;
    }
}

Playground link

Side note: I've left construction in the above in case it's meant to be a method, but suspect you meant constructor.