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Finding literals rather than patterns in Lua

Lua's string.find looks for a PATTERN ... it's not as big an sophisticated as a regular expression match (but then it's got nothing like as big a footprint!), but neither is it a literal match. So if I want to search a string for - literally - a ".", I need to protect it. I can do this either with a % character, or by providing an extra true parameter to my find, meaning "I really want literals"

Let's look for a dot:

pos,posend = string.find(sed,".",at,true)
if not pos then break end


string.find returns both the start and the end character numbers of the match; in the case of a single literal character match such as mine here, the two values will be identical and indeed I could have left out ,posend save for my desire to provide an illustration.

Complete example source here, courses on the subject here.
(written 2009-08-11)

 
Associated topics are indexed as below, or enter http://melksh.am/nnnn for individual articles
U104 - Lua - Strings and the String library.
  [4274] String formatting in Lua - string.format as a wrapper for sprintf - (2014-05-07)
  [2500] Dynamically formatting your results (Lua) - (2009-11-10)
  [2357] Checking if the user has entered a number (Lua) - (2009-08-13)
  [1744] Lua examples, Lua Courses - (2008-08-08)


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Finding literals rather than patterns in Lua
Dot, dot, dot in Lua - variable length parameter lists
The indexed and hashed parts of a Lua table
Automatically initialising Lua variables the first time
Reading a data file and sensing EOF in Lua
World Flags in your PHP pages
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