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On 2017-10-10 04:13 PM, Dirk Laurie wrote:

function end_of_string_literal(word,quote)
   local m,pos
   repeat
     m, pos = string.match(word, "(\\*)"..quote.."()", pos or 2)
   until m == nil or #m % 2 == 0
   return pos
end

end_of_string_literal("abc'defgh'ijkl","'")
5

Should not the answer rather be 11?

And would not

function end_of_string_literal(word,quote)
   local pattern = '()%b' .. quote:rep(2) ..'()'
   return select(2,word:match(pattern))
end

be somewhat simpler?


It says "end". It implies you already know the start.

"\""





2017-10-10 20:25 GMT+02:00 Soni L. <fakedme@gmail.com>:


On 2017-10-10 02:44 PM, Dirk Laurie wrote:

2017-10-10 14:52 GMT+02:00 Soni L. <fakedme@gmail.com>:

local m, pos
repeat -- TODO fix
    m, pos = string.match(word_eol[2],
"%f[\\"..cw:sub(1,1).."](\\*)"..cw:sub(1,1).."()", pos)
until #m % 2 == 0

I'm having issues making this work.

The title of the post describes a trivial problem (answer: #str+1).

I can't work by reading it what problem the code tries to solve.


Sorry, I meant end of string literal.

After a few hours I came up with this, which works:

local m, pos
repeat
   m, pos = string.match(word_eol[2], "(\\*)"..cw:sub(1,1).."()", pos or 2)
until m == nil or #m % 2 == 0

(cw:sub(1,1) being one of " or ')


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