Question: How do I create an instance of a BigIntString?

[Industrial]

I have this in my Schema:

  minimumReward: S.BigIntFromString,

and get this error:

The types of 'input.minimumReward' are incompatible between these types.
    Type 'string' is not assignable to type 'BigIntString'.
      Type 'string' is not assignable to type 'Tagged<BigIntStringBrand>'.ts(2322)

[jacob-alford]

There are a few of ways to do this.

• Coerce a string you know to be a BigIntString using fp-ts 'unsafeCoerce'. I use this a lot at work when I know what I'm coercing abides the underlying branded type.
• Cast the type as the right type. Most of the time casting number to Int, number to Float, Date to SafeDate and string to BigIntString is castable with one as clause.
• Use const guard = deriveInputGuard(S.BigIntString) and use guard.is() to instruct typescript that the type is safely the underlying type.

Most of the time for branded types I know to be valid values, for example, Int's, Floats, SafeDate's, etc I just use unsafeCoerce because there is no concern about getting it wrong. If the source of the type is unknown, i.e. from a network request then you should first .decode() from deriveTranscoder() or .is() from deriveGuard() / deriveInputGuard() to ensure that the type matches the expected type.

[jacob-alford]

Also, to clarify, casting and using unsafeCoerce should only be done when you write the value of the type yourself (or as an escape hatch when you're certain the type coming in is trustworthy).

If the value is coming from a somewhere else, you should lift the function into an Either type and use Transcoders to decode the value first. This will guarantee type safety. And alternatively you can use guard in this context for the same effect.

[Industrial]

Okay, thanks.