what is apostrophe type in scheme

lisp, scheme

Solution

Notice that the `cond` expression is not doing what you think. What's really happening is this:

(cond ((member '1 '(some-function))
       (display #t))
      (else
       (display #f)))

In other words: the number `1` is being quoted and the expression `'(some-function)` is being interpreted as a single-element list with the symbol `some-function` as its only member. Regarding the first question, this expression:

'1'

… is invalid in Scheme - try typing it in the evaluation window, nothing will happen: the first quote applies to the number `1`, and the second quote is expecting further input, so anything that's written after it will get quoted. FYI double quotes mean a string, as in many other languages: `"1"`. But a single quote indicates a quoted expression, that evaluates to itself:

'1
=> 1

And it's just shorthand for this:

(quote 1)
=> 1

Which in the above expression is unnecessary, a number already evaluates to itself:

1
=> 1

Now, about the second question, it doesn't make sense because `'1'` is not a type, as explained above.

Problem

I have condition that uses the member function: ``` (cond ((member '1' (some-function)) (display #t)) (else (display #f))) ``` it works fine but i still couldn't find the answers to: 1)what is the type of '1'? 2)i have the next expression (lambda(x)(= x 1)) how can I convert to the same type of '1'?

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