John Cowan <[email protected]> wrote:
> ... > Someone asked about what `apply` does with fexprs. In classic Lisps, > fexprs don't know anything about where their operands come from any > more > than other procedures do, so the environment at the point of > application > is used to interpret variables, there being in fact no other. In > Kernel, > it is a domain error to invoke `apply` on a fexpr. But I think the latter is (also) because with `apply', the arguments to the function being applied have already been evaluated; yet another reason why fexprs (never mind macros) are not really like functions. ---Vassil. -- Would you like your metaphors shaken or stirred? Vassil Nikolov | Васил Николов | <[email protected]> _______________________________________________ Scheme-reports mailing list [email protected] http://lists.scheme-reports.org/cgi-bin/mailman/listinfo/scheme-reports
