What is the best way to assign a type property of an argument to a generic in TypeScript? Here's the code snippet:
const foo = <T = someObject.bar>(someObject: {[string]: any}): T => {
return someObject.bar
}
How can we set the type of T to be equal to the value of the bar property in the object someObject?
Are you looking to ensure that the argument includes the property bar?
interface WithBar<T> {
bar: T
}
You can implement your function like this:
const foo = <T>(someObject: WithBar<T>): T => {
return someObject.bar
}
If you prefer not to define the WithBar type, you can simplify it to:
const foo = <T>(someObject: { bar: T }): T => {
return someObject.bar
}
This function can be used with various types:
const barFoo1 = { bar: "hello", foo: "world"};
const myBar1: string = foo(barFoo1);
const barFoo2 = { bar: 123, foo: 456};
const myBar2: number = foo(barFoo2);
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