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Parse, don't validate (2019)

(lexi-lambda.github.io)
398 points declanhaigh | 1 comments | 07 Mar 23 08:47 UTC | HN request time: 0.21s | source
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bruce343434 ◴[07 Mar 23 10:54 UTC] No.35053912[source]▶
Note that this basically requires your language to have ergonomic support for sum types, immutable "data classes", pattern matching.

The point is to parse the input into a structure which always upholds the predicates you care about so you don't end up continuously defensively programming in ifs and asserts.

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mtlynch ◴[07 Mar 23 11:13 UTC] No.35054046[source]▶
I get a lot of value from this rule even without those language features.

I follow "Parse, Don't Validate" consistently in Go. For example, if I need to parse a JSON payload from an end-user for Foo, I define a struct called FooRequest, and I have exactly one function that creates a FooRequest instance, given a JSON stream.

Anywhere else in my application, if I have a FooRequest instance, I know that it's validated and well-formed because it had to have come from my FooRequest parsing function. I don't need sum types or any special language features beyond typing.

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jotaen ◴[07 Mar 23 11:28 UTC] No.35054157[source]▶
My main take-away is the same, I wonder though whether “parse, don’t validate” is the right term for it. To me, “parse, don’t validate” somehow suggests that you should do parsing instead of validation, but the real point for me is that I still validate (as before), plus I “capture”/preserve validation success by means of a type.
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1. lkitching ◴[07 Mar 23 12:01 UTC] No.35054377[source]▶
The post is suggesting that parsing and validation are different things, since the output of a parser captures the properties being checked in the type, and validation does not. Downstream consumers of validated input cannot rely on the properties that were validated since the representation type doesn't encode them e.g. the non-emptiness of a list.