Re: pattern matching problem

[Coda Highland <chighland@...>]

On Fri, Aug 30, 2013 at 1:52 PM, Choonster TheMage
<choonster.2010@gmail.com> wrote:
> On 31 August 2013 06:42, Coda Highland <chighland@gmail.com> wrote:
>> On Fri, Aug 30, 2013 at 1:29 PM, Choonster TheMage
>> <choonster.2010@gmail.com> wrote:
>>> On 31 August 2013 06:14, Geoff Smith <spammealot1@live.co.uk> wrote:
>>>> Hi
>>>>
>>>> I always seem to have difficulty with pattern matching in Lua, here is
>>>> another example that has stumped me so far
>>>>
>>>> I am trying to trim a number string to remove non significant trailing
>>>> zeros. So for example
>>>>
>>>> "123.450000000" = "123.45"
>>>> "1234500" = "1234500"
>>>> "12345.00" = "12345.0"
>>>>
>>>> I could of course write a function that iterates through the string from the
>>>> back dropping zeros, but I thought there is probably a more elegant solution
>>>> using Lua pattern matching functions.
>>>>
>>>> Can this be done neatly with pattern matching ? Thanks for any suggested
>>>> code snippets
>>>>
>>>> Regards Geoff
>>>>
>>>>
>>>>
>>>
>>> Try this pattern:
>>>
>>>     local stripped = fullNumber:match("^(%d*%.?0?%d-)0*$")
>>>
>>> Where fullNumber is the string containing the number that you want to
>>> remove trailing zeroes from.
>>>
>>
>> I have a hesitation on that solution -- if there isn't a decimal
>> point, that's still going to truncate trailing zeroes, which is
>> clearly incorrect behavior.
>>
>> /s/ Adam
>>
>
> It doesn't do that under Lua 5.1.5, 5.2.1 or LuaJIT 2.0.1
>
> Running this code:
>
> local function trim0( s )
>     return s:match("^(%d*%.?0?%d-)0*$")
> end
>
> print( trim0 "123.450000000" )
> print( trim0 "1234500" )
> print( trim0 "12345.0" )
>
> Gives this output:
>
> 123.45
> 1234500
> 12345.0
>
> The first %d* should capture all digits when there's no decimal point,
> since it's a greedy pattern item.
>

... bah! You're right. I should have run it before saying anything. I
misread the pattern. Sorry for the noise.

/s/ Adam