4 ms·
It is not. I made a preprocessor which expands "lambda literals" into lambdas. [|y| (cons x y)] turns into (lambda (y) (cons x y))
by billsix 9y ago
It is not. I made a preprocessor which expands "lambda literals" into lambdas.
[|y| (cons x y)] turns into (lambda (y) (cons x y))
- agumonkey 9y agoThanks