Assigning a list of atoms in Scheme

Question

I'm trying to learn Scheme from the book "The Little Schemer" using DrScheme on a Macintosh. It starts with things like "What is the car of l where l is the argument (a b c)?"

I understand that the answer to this question is a, but I'm not able to actually figure out what to type into Dr Scheme to "follow along". A simple idea on how to assign to l such as > (def l ('a 'b 'c))

gives me the error: "function call: expected a defined name or a primitive operation name after an open parenthesis, but found something else"

Answer 1

Try this:

(define l '(a b c))

Here, the quote applies to the whole list (a b c).

Answer 2

Or, just use

(car '(a b c))

What the error is telling you is this: when the reader sees a list, it wants to treat the first element of the list as a function. That's how (+ 1 2 3) works: it invokes hthe function +. (Strictly, it looks at the symbol '+ and finds that there is a function bound to that, then invokes that function.)

So, when you type

(define l ('a 'b 'c))

it looks at the inner list and then wants to find a function. Instead it find s the symbol named a. Since there isn't a function there, you get the error. If, instead, you type

(define l '(a b c))

you've told the reader via the quote that it's to treat that as a list without trying to interpret it.

Here's an example from MIT Scheme on a Mac:

1 ]=> (define li '(a b c))

;Value: li

1 ]=> (car li)

;Value: a

1 ]=> (car '(a b c ))

;Value: a

Answer 3

To define something:

(define <name> <value>)

So to define l:

(define l '(a b c))

This defines l as the list (a b c). The single quote mark quotes whatever is after it, whether it's a symbol or a list, which means it's not evaluated but read as-is. You don't, however, quote the name that you're setting it to. Thankfully, this is one of the very small number of operations in Scheme where you have this inconsistency.

Answer 4

Non-abbraveted form of

(define l '(a b c))

is

(define l (quote (a b c)))

or in this case even this:

(define l (list 'a 'b 'c))

without a "list" part it tries to execute 'a, obviously