Read input into string in Lisp reader macro

Question

I am trying to make a reader macro that would convert @this into "this". This is what I currently have:

(defun string-reader (stream char)
   (declare (ignore char))
   (format nil "\"~a\"" (read-line stream t nil t))   
)    
(set-macro-character #\@ #'string-reader )

The problem is that this requires that I put a newline after ever @this. I've also tried it with (read), but that just returns the variable test, which has not been set. I can't just hard-code the number of characters after the @ symbol, because I don't know how many there would be. Is there any way to fix this?

Edit: is the only way to do this to loop over read-char and peek-char, reading until I get to #),#\space, or #\Newline?

Answer 1

You can try to use read and then look at what it returns:

(defun string-reader (stream char)
   (declare (ignore char))
   (let ((this (let ((*readtable* (copy-readtable)))
                 (setf (readtable-case *readtable*) :preserve)
                 (read stream t nil t))))
     (etypecase this
       (string this)
       (symbol (symbol-name this)))))

(set-macro-character #\@ #'string-reader)

Above would allow @This and @"This", but not @333.

This version just reads a string until whitespace:

(defun read-as-string-until-whitespace (stream)
  (with-output-to-string (out-stream)
    (loop for next = (peek-char nil stream t nil t)
          until (member next '(#\space #\newline #\tab))
          do (write-char (read-char stream t nil t) out-stream))))

(defun string-reader (stream char)
   (declare (ignore char))
   (read-as-string-until-whitespace stream))

(set-macro-character #\@ #'string-reader)

Example:

CL-USER 21 > @this
"this"

CL-USER 22 > @42
"42"

CL-USER 23 > @FooBar
"FooBar"