Typescript create uber-strict types

Question

If I create a custom type such as:

type LowercaseString = string

And then use it in a function like this:

function displayLowercaseString(input: LowercaseString) {
  ...
}

displayLowercaseString('UPPERCASE')

The above example compiles perfectly and runs, since 'UPPERCASE' is of type string and so is LowercaseString

My question is, is it possible to have a TS error such as type string does not match LowercaseString so that I am forced to put every string through a function like this before using it in displayLowercaseString:

function stringToLowercase(input: string): LowercaseString {
  return input.toLowercase as LowercaseString
}

What you're looking for is essentially github.com/microsoft/TypeScript/issues/41160, which has been open for years and is not likely to be resolved anytime soon. — Etheryte
– Etheryte, Commented Jul 10, 2022 at 14:18
By using a string in my example I think I have created a misleading question. I don't want any kind of validation of the content of the strings. I effectively just want type names to be matched. type ID = number is another example, where I wouldn't want a function that accepts type ID to just accept any number. It should only accept variables with the type ID — StanFlint
– StanFlint, Commented Jul 10, 2022 at 15:16

Matthieu Riegler · Accepted Answer · 2022-07-10 15:24:12Z

1

You can use a branded type :

type LowercaseString = string & { __brand: 'lower' };


function stringToLowercase(input: string): LowercaseString {
  return input.toLowerCase() as LowercaseString
}

function displayLowercaseString(input: LowercaseString) {
}

const uppercaseString = 'UPPERCASE';
displayLowercaseString(uppercaseString) // nope 
displayLowercaseString(stringToLowercase(uppercaseString)) // ok

Playground

answered Jul 10, 2022 at 15:24

Matthieu Riegler

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2 Comments

StanFlint Over a year ago

Thank you very much, this is exactly what I was looking for and works perfectly! Am I correct in assuming that this does not add any additional code when compiled to JavaScript? (as with normal TS types)

Matthieu Riegler Over a year ago

@StanFlint Yes, this is purely typechecking code. You can look at the generated code in the playground !

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