How to resolve the algorithm Arithmetic evaluation step by step in the Lua programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Arithmetic evaluation step by step in the Lua programming language
Table of Contents
Problem Statement
For those who don't remember, mathematical precedence is as follows:
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Arithmetic evaluation step by step in the Lua programming language
Source code in the lua programming language
require"lpeg"
P, R, C, S, V = lpeg.P, lpeg.R, lpeg.C, lpeg.S, lpeg.V
--matches arithmetic expressions and returns a syntax tree
expression = P{"expr";
ws = P" "^0,
number = C(R"09"^1) * V"ws",
lp = "(" * V"ws",
rp = ")" * V"ws",
sym = C(S"+-*/") * V"ws",
more = (V"sym" * V"expr")^0,
expr = V"number" * V"more" + V"lp" * lpeg.Ct(V"expr" * V"more") * V"rp" * V"more"}
--evaluates a tree
function eval(expr)
--empty
if type(expr) == "string" or type(expr) == "number" then return expr + 0 end
--arithmetic functions
tb = {["+"] = function(a,b) return eval(a) + eval(b) end,
["-"] = function(a,b) return eval(a) - eval(b) end,
["*"] = function(a,b) return eval(a) * eval(b) end,
["/"] = function(a,b) return eval(a) / eval(b) end}
--you could add ^ or other operators to this pretty easily
for i, v in ipairs{"*/", "+-"} do
for s, u in ipairs(expr) do
local k = type(u) == "string" and C(S(v)):match(u)
if k then
expr[s-1] = tb[k](expr[s-1],expr[s+1])
table.remove(expr, s)
table.remove(expr, s)
end
end
end
return expr[1]
end
print(eval{expression:match(io.read())})
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