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Lazy evaluation is the default in Haskell, but that doesn't mean strict evaluation is unavailable. Here's an example of a type definition that specifies strict

Lazy evaluation is the default in Haskell, but that doesn't mean strict evaluation is unavailable. Here's an example of a type definition that specifies strict evaluation https://github.com/vincenthz/hs-tls/blob/master/core/Network/TLS/Record/State.hs#L39 https://github.com/vincenthz/hs-tls/blob/master/core/Network... See? They don't have to switch languages after all.

amalcon 13y ago
This is an excellent point, and I didn't mean to imply that it's impossible to implement TLS securely in Haskell (though my original comment did seem to; I plan to fix that shortly). Only that there are more natural choices that give the same primary advantage (a strong algebraic type system).