How do I pass a list as a list of arguments in racket?

list parameters scheme racket
peskal picture peskal · Aug 10, 2011 · Viewed 7.3k times · Source

I have a statement like this:

 ((lambda (a b c) (+ a b c)) 1 2 3) ; Gives 6

And I would like to be able to also pass it a list as so:

((lambda (a b c) (+ a b c)) (list 1 2 3))

...except this doesn't work because the entire list is passed as 'a.' Is there is a way to explode the list into arguments?

What I'm looking for is something similar to the * character in Python. For those of you unfamiliar with the syntax:

 def sumthree(a, b, c):
   print a + b + c

 sumthree(1, 2, 3) # Prints 6
 sumthree(*(1, 2, 3)) # Also prints 6

Answer

Chris Jester-Young picture Chris Jester-Young · Aug 11, 2011

That operation is called apply.

(apply + (list 1 2 3))   ; => 6

apply "expands" the last argument; any previous arguments are passed as is. So these are all the same:

(apply + 1 2 3 (list 4 5 6))
(apply + (list 1 2 3 4 5 6))
(+ 1 2 3 4 5 6)

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