Why is the Keyword `as` Needed When Narrowing TypeScript Interfaces With an `if`?

Why is the Keyword `as` Needed When Narrowing TypeScript Interfaces With an `if`?

Jun 26 '21 Comments: 1 Answers: 2

1

Given some interfaces for Circle and Square as follows:

export default interface Circle {
  radius: number
}

export default interface Square {
  sideLength: number;
}

I can use as to calculate the area properly:

function getArea(shape: Circle | Square) {
  if ((shape as Circle).radius) {
    return Math.PI * (shape as

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