Deciding the conditional type within an IDE while using an If statement

When dealing with a type that can have two formats based on the value of one of its keys, such as:

type singleOrMultiValue = {isSingle: true, value: string} | {isSingle: false, set: Array<string>}

I have found it useful in preventing errors like

const val : singleOrMultipleValue = {isSingle: false, value: '1'}
.

However, when trying to determine the type of a variable at runtime, I came across an issue:

if (val.isSingle) {
   console.log(val.value)
} else {
   console.log(val.set[0])
}

This resulted in an error stating that value does not exist on type singleOrMultiValue for {isSingle: false}. Is there a way to specify within this if scope that isSingle has a value of true? Or do I need to explicitly declare another variable or use as any?

Any insights would be greatly appreciated!

Answer №1

Discovered the resolution - I learned that VS Code is able to identify the value within an if statement when written as else if (val.isSingle == false), rather than using else if (!val.isSingle) or else {.

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