use Perl

• List nit (Score:1)

  by runrig (3385)

  "list" is already a function name, so I wouldn't use it as a variable. This wouldn't happen if the language had sigils...
  • Re:List nit (Score:1)

    by jrockway (7171) on 2008.01.31 17:08 (#60780) Journal

    This isn't a problem. Symbols get their values from different "slots" depending on context. When you write (list 'foo 'bar), list is called as a function. When you write (setq list 42) and then (+ list list), you get 84. It's very clear. The only problem is calling a function saved in the value slot of a symbol: (defun foo (a b) (- a b)) (let ((foo (lambda (a b) (+ a b)))) (foo 2 1) ; 1 (funcall foo 2 1)) ; 3 Oh well, you can't have everything.

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    Parent

    • Re: (Score:1)

      by jrockway (7171)

      Ugh, apologies for the worthless formatting.
    • Re: (Score:1)

      by chromatic (983)

      This isn't a problem.

      ... not for the compiler anyway. Some of the rest of us like to optimize for the slower parts of the process, often known as wetware.
      • Re: (Score:1)

        by jrockway (7171)

        FWIW, Perl suffers from this problem, sort of:
        
        sub foo {}
        my $foo = sub {};
        foo();
        $foo->();
        my $foo = "hello";
        $foo->(); # death
        
        Anyway, lisp is what it is.  It's straightforward to write a macro such that:
        
        (with-sigils (&foo) (foo $foo))
        
        expands to
        
        (progn (funcall foo) (foo foo)))
        
        If you really care I'll try it out and blog the code :)
    • Re: (Score:1)

      by runrig (3385)

      Disclaimer: the last Lisp I used was AutoLisp...many years ago. I don't remember if it behaved the way you describe or not. I later (after posting) figured this wouldn't be a problem in this instance anyway because "list" was (dynamically?) scoped to the function. But I still don't think I'd call a variable "list" in the language (w/o sigils) :-)