Re: Smallest Lua-only single-file JSON parser

[Patrick Rapin <toupie300@...>]

As an exercise, I tried to wrote the really minimum JSON parser.
It is below (only 10 lines of code)...
It first converts the JSON snipped into a valid Lua expression using
string patterns, then evaluates it.
Bugs apart, the only missing feature is the \uXXXX string escape.

function decode_json(json)
	local str = {}
	local escapes = { r='\r', n='\n', b='\b', f='\f', t='\t', Q='"',
['\\'] = '\\', ['/']='/' }
	json = json:gsub('([^\\])\\"', '%1\\Q'):gsub('"(.-)"', function(s)
		str[#str+1] = s:gsub("\\(.)", function(c) return escapes[c] end)
		return "$"..#str
	end):gsub("%s", ""):gsub("%[","{"):gsub("%]","}"):gsub("null", "nil")
	json = json:gsub("(%$%d+):", "[%1]="):gsub("%$(%d+)", function(s)
		return ("%q"):format(str[tonumber(s)])
	end)
	return assert(loadstring("return "..json))()
end