Re: boolean operators

[Rici Lake <lua@...>]

On 28-Sep-06, at 5:53 PM, Glenn Maynard wrote:

> On Thu, Sep 28, 2006 at 10:46:58PM +0000, David Given wrote:
> > arg = "5"
> > print (arg+0)
> >
> > Likewise, some compiler sugar to convert to treat '+a' as '0+a' would 
> > be
> > easy, not involve any additional opcodes, and would optimise away to
> > nothing in most cases.
>
> echo 'for i = 1,10000000 do local a = 5*(i); end' | time luac -l -
> echo 'for i = 1,10000000 do local a = 5*(0+i); end' | time luac -l -
>
> Lua's compiler doesn't do much (any?) optimization.

It could only optimize that by recognizing that i was a number; 
otherwise, 0+i could trigger an arbitrary metamethod. (The same comment 
applies to - -i). Here I think that use of tonumber() is better, 
particularly since tonumber can be used as a test:

i = tonumber(s)
if i then
  -- at this point i is known to be a number
end

If I understood your desire for "unary +" correctly, your needs would 
be satisfied if the lexer recognized + as the first character in a 
numeric lexeme, no? The use case did not seem to require the use of + 
as an operator.