[kazinator]

From your brief description that is likely incomplete, it looks as if the length? function is treated as a prefix operator of low precedence relative to the infix operators. The infix operators are all at the same precedence level and have left-to-right associativity.

I made an infix parser in which certain prefix operators (named math functions) have a low precedence. This allows for things like

  1> log10 5 + 5   ;; i.e. log10 10
  1.0

But a different prefix operator, like unary minus, binds tighter:

  2> - 5 + 5
  0

I invented a dynamic precedence extension to Shunting Yard which allows this parse:

  3> log10 5 + 5 + log10 5 + 5    ;; i.e. (log10 5 + 5) + (log10 5 + 5)
  2.0

Functions not registered with the parser are subject to a phony infix treatment if their arguments look like they might be infix and thus something similar happens to your Red example:

  4> len "123" - 2
  ** -: invalid operands "123" 2

"123" - 2 turns into a single argument to len, which does not participate in the infix parsing at all. log10 does participate because it is formally registered as a prefix operator.

The following are also the result of the "phony infix" hack:

  4> 1 cons 2
  (1 . 2)
  5> 1 cons 2 + 3
  (1 . 5)

Non-function in first place, function in second place leads to a swap: plus the arguments are analyzed for infix.

[sph]

Not sure why you got downvoted, but Rebol is not a Lisp. It doesn't work because of precedence rules or special rules, but because arguments accumulate until there are enough to eval the previous function in the stack, so you can do stuff like

  print tostring 5 + cos pi

Works a bit like a shift/reduce parser, with heavy use of fexprs (blocks in Rebol parlance)

I know you enjoy Lisps, so you might like this toy Rebol evaluator written in Scheme: http://ll1.ai.mit.edu/marshall.html

[Izkata]

I'm guessing because their understanding is still wrong:

  3> log10 5 + 5 + log10 5 + 5    ;; i.e. (log10 5 + 5) + (log10 5 + 5)
  2.0

In Rebol this would be equivalent to (log10 (5 + (5 + (log10 (5 + 5)))))