2 ms·
Internally, Haskell's intermediate representation is a version of the lambda calculus. Which would mean that, practically speaking, Haskell is largely just one
by mumblemumble 6y ago
Internally, Haskell's intermediate representation is a version of the lambda calculus. Which would mean that, practically speaking, Haskell is largely just one big pile of closures.
Which really shouldn't be a surprise. After all, you can't curry if you can't close.