Patterns

How do I make success data available only in the success branch and errors only in the failure branch?

Give every variant one shared literal field. Checking it narrows the whole object, not just the field.

The recipe

type Result<T> =
  | { status: 'success'; data: T }
  | { status: 'failure'; error: Error }

function unwrap<T>(result: Result<T>): T {
  if (result.status === 'success') {
    return result.data
  }
  throw result.error
}

The build compiles this and checks each result below.

How it works

01
```
status: 'success'
```
A literal discriminant identifies this member unambiguously.
02
```
if (result.status === 'success')
```
Control-flow analysis now knows every other property on the selected variant.

What you get

```
Result<number>['status']
```
→"success" | "failure"

Extract<Result<number>, { status: 'success' }>

→{ status: "success"; data: number; }

```
ReturnType<typeof unwrap<string>>
```
→string

Where it goes wrong

Optional fields on one broad object do not create the same correlation. `status: "success" | "failure"` beside optional `data` and `error` still lets impossible combinations through.

Takeaway

Use a union of complete states, not one object full of optional state fragments.