Well, # is the length operator and it’s meant to return the length of
whatever follows, provided what follows has a concept of length/size (e.g.,
table, string), so one -- I being that one, I guess :) -- would expect if a
number follows it would return its length (of its common printable
representation), rather than give error. In other words, I would expect to
#number to be equivalent to #(#number..’’)
In any case, in a ‘loosely typed language’ giving an error when there is a
non-error alternative behavior does not serve any useful purpose, does it?
Obviously, it’s not a big issue as it’s easy to overcome, but it seems (to
me, at least) not the best possible behavior for #number
(Another thing that I would expect is the IN operator to work like in
Python when used with strings. But that’s another discussion...)
Sent: Friday, November 21, 2014 9:04 PM
Subject: Re: ## operator (error)
I didn’t know you could do ##, but then again I didn’t think about it
until you mentioned it. I actually use the length of a number (as
string) quite often.
So, being shorter than what I currently do, I tried ## and I have to ask
(not you necessarily, whoever knows), why an error?
x = '1234567890'
print(#x)
print(#(#x..'')) --
this works (prints 2, the length of 10 as
string)
print(##x) --
this fails "attempt to get length of a number value"
If numbers and strings can be mixed in Lua (as
in the example that works above), why is ##a an error if x is a number, and
Lua does not implicitly convert the number to a string to have its length
taken, rather than give an error? The length of a number is a meaningful
thing. Of course one could say it should return the length as the number
of bytes required to stored the number but this is the less likely
interpretation one would expect.
TIA
Sent: Friday, November 21, 2014 5:37 PM
Subject: Re: [Proposal] ## operator
Probably a scenario which is totally unlikely and broken on so many other
aspects.. but this code
t = { 1, 2, 3 };
setmetatable(t, { __len =
function(tbl) return tostring(rawlen(tbl)); end });
x = ##t;
-- x = 1
actually works and would break if ## would be
implemented. As I said, totally unlikely and super-dirty to starts of,
but at the moment the parser treats (correctly) ## as two executions of the #
operator and if a __len meta returns a string or table, it's actually
"working" code (for some definition of working). --
Marco
Because # is not a string
method. --
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