转换(条件？然后: else)到带有lpeg的ifthenelse(cond，then，else)

Question

我正在尝试使用lpeg lua解析器将像'a?(b?c:d):e'这样的字符串转换为另一个字符串'ifthenelse(a,ifthenelse(b,c,d),e)'。我正在慢慢地学习如何使用lpeg，但我仍然找不到一个合适的解决方案来实现这一点。有什么想法吗？

以下是我到目前为止所做的工作。

local lpeg = require("lpeg")

local S, P, R = lpeg.S, lpeg.P, lpeg.R
local C, Cc, Ct = lpeg.C, lpeg.Cc, lpeg.Ct
local Cf, Cg, Cs = lpeg.Cf, lpeg.Cg, lpeg.Cs
local V = lpeg.V

local thenop = P("?")
local elseop = P(":")
local openpar = P("(")
local closepar = P(")")
local digit = R("09")
local letter = R("az") + R("AZ")

local parser = 
   P({
    "F",
    F = V("E") * (thenop * V("E") * elseop * V("E"))^0,
    E = (letter + digit)^1 + (openpar * V("F") * closepar)
 }) -- * -1 -- Is it needed?
print(lpeg.match(parser,"a?(b?c:d):e"))
print(lpeg.match(parser,"a"))

Answer 1

这是William Ahern在lua列表上给出的另一个解决方案。

local lpeg = require("lpeg")

lpeg.locale(lpeg)

local function tr(a, b, c)
   if not b then
       return a
   else
       return string.format("ifthenelse(%s,%s,%s)", a, b, c)
   end
end

local var = lpeg.C(lpeg.alpha * (lpeg.alnum^0))

local E, G = lpeg.V"E", lpeg.V"G"

local grammar = lpeg.P{ "E",
   E = ((var + G) * (lpeg.P"?" * E * lpeg.P":" * E)^-1) / tr,
   G = lpeg.P"(" * E * lpeg.P")",
}

print(lpeg.match(grammar, "a?(b?c:d):e"))

Answer 2

我稍微改变了一下语法，但我想还是一样的：

S = E "?" E ":" E | E
E = "(" S ")" | T
T = (L | D) +
L = [a-z] | [A-Z]
D = [0-9]

不带捕获的语法：

local grammar = lpeg.P{
  "S";
  S = (lpeg.V"E" * lpeg.P"?" * lpeg.V"E" * lpeg.P":" * lpeg.V"E") + lpeg.V"E",
  E = (lpeg.P"(" * lpeg.V"S" * lpeg.P")") + lpeg.V"T",
  T = (lpeg.V"L" + lpeg.V"D")^1,
  L = lpeg.R("az") + lpeg.R("AZ"),
  D = lpeg.R("09")
}

带有捕获的语法：

local grammar2 = lpeg.P{
  "S";
  S = (lpeg.Cs(lpeg.V"E") / "ifthenelse(%1") * (lpeg.P"?" / ",") * lpeg.V"E" * (lpeg.P":" / ",") * (lpeg.Cs(lpeg.V"E") / "%1)") + lpeg.V"E",
  E = (lpeg.P"(" / "") * lpeg.V"S" * (lpeg.P")" / "") + lpeg.V"T",
  T = (lpeg.V"L" + lpeg.V"D")^1,
  L = lpeg.R("az") + lpeg.R("AZ"),
  D = lpeg.R("09")
}

捕获替换：

具有空字符串的变量“(”和")“(变量和lpeg.P")" / ""))

• First变量与”ifthenelse(“(lpeg.Cs(lpeg.V"E") / "ifthenelse(%1"))
• "?”“
• ”匹配和":“与"，”(变量和lpeg.P":" / ",")
• Last变量与“(lpeg.Cs(lpeg.V"E") / "%1)")

”匹配)“lpeg.P"?" / ","”

一些随机测试(在注释中输出)：

print( lpeg.match( lpeg.Cs(grammar2), "a") )
-- a
print( lpeg.match( lpeg.Cs(grammar2), "a?b:c") )
-- ifthenelse(a,b,c)
print( lpeg.match( lpeg.Cs(grammar2), "a?(i?j:k):c") )
-- ifthenelse(a,ifthenelse(i,j,k),c)
print( lpeg.match( lpeg.Cs(grammar2), "(a?(i?j:(x?y:z)):b)?c:(u?v:w)") )
-- ifthenelse(ifthenelse(a,ifthenelse(i,j,ifthenelse(x,y,z)),b),c,ifthenelse(u,v,w))

我希望你能从这里继续下去。