Standing alone, an argument can never be fully validated without

Recently, while delving into the valuable resource titled Effective TypeScript by Dan Vanderkam, I stumbled across an intriguing scenario that left me puzzled. Within a code snippet presented in the book, there was a line - shape; that seemed perplexing to me. Despite appearing straightforward at first glance, it failed to resonate with my understanding. What is the significance of this shape; statement? Moreover, why is it repeated in the code? Could this possibly be a typo?

interface Square {
  kind: 'square';
  width: number;
}
interface Rectangle {
  kind: 'rectangle';
  height: number;
  width: number;
}
type Shape = Square | Rectangle;

function calculateArea(shape: Shape) {
  if (shape.kind === 'rectangle') {
    shape;  // I find this confusing!
    return shape.width * shape.height;
  } else {
    shape;  // Puzzling, isn't it?
    return shape.width * shape.width;
  }
}

Answer №1

Oops, I forgot to check the book's introduction. The author mentions:

At times, I may include no-op statements to highlight the type of a variable on a specific line of code:

function bar(y: number | number[]) {
  if (Array.isArray(y)) {
    y; // Type is number[]
  } else {
    y; // Type is number
  }
}

The y; lines serve solely as examples of the types in each branch of the conditional. It's not necessary (and actually discouraged) to have such statements in your own code.

Answer №2

It seems like there may have been a typo in the code. The presence of this shape element doesn't seem to be causing any issues, so removing it should not cause any problems.

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