Re: Why is booleanp defined this way?

[Pascal J. Bourguignon]

Emanuel Berg <embe8573@student.uu.se> writes:

> The "normalization" of which you speak should rather
> look something like this:
>
>     (defun normalize-boolean (obj)
>       (if obj t) ) ; implicit (if obj t nil)
>
>     (normalize-boolean 1)   ; t
>     (normalize-boolean nil) ; nil

You may want to compare:

    (defun normalize-boolean (obj)
      (if obj t))
    (disassemble (byte-compile 'normalize-boolean))
    byte code:
      args: (obj)
    0       varref    obj
    1       goto-if-nil-else-pop 1
    4       constant  t
    5:1     return    

with:

    (defun g (x) (not (not x)))
    byte code:
      args: (x)
    0       varref    x
    1       not       
    2       not       
    3       return    

    (disassemble (byte-compile 'f))


> So I think `booleanp' shouldn't be thought of as
> a normalizer but rather as a type predicate, much like
> them `stringp', `integerp', and so on.

Of course.  That's what the "p" in "booleanp" means!


-- 
__Pascal Bourguignon__                 http://www.informatimago.com/
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