CODEWITHSUNDEEP
scheme
briskBuilding
briskBuilding

Reputation: 21

Is there a limit to how many car and cdr we can chain in Scheme?

Some answers on here just say that you can chain multiple car and cdr to reduce the clutter they create so, you could use (cadr (list)) instead of (car (cdr (list) )), but I have tried chaining multiple car and cdr and sometimes it will stop working on like the 5th one or something, I tried googling if there is a limit or something but couldnt find any answers.

specifically my problem shows here -> trying to get 7 from a list only using car and cdr.

list is (1(2(3(4(5(6(7)))))))))

(display (car(car(cdr(car(cdr(car(cdr(car(cdr(car (cdr (car (cdr '(1(2(3(4(5(6(7)))))))))))))))))))))

-> shows 7

(display (caadadadadadadr '(1(2(3(4(5(6(7)))))))))

-> shows error

Is this not how this works?

Upvotes: 2

Views: 192

Answers (1)

amalloy
amalloy

Reputation: 91857

You aren't specific about which Scheme you're using. But for every Scheme implementation I'm aware of, there's a limit to how many car/cdr compositions are provided. In R5RS, for example, we see

library procedure: cddddr pair

These procedures are compositions of car and cdr, where for example caddr could be defined by

(define caddr (lambda (x) (car (cdr (cdr x))))).

Arbitrary compositions, up to four deep, are provided. There are twenty-eight of these procedures in all.

Upvotes: 1

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