Desugar Lua operators
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28-10-2019 - |
Question
Since Lua supports first-class functions, I'd like to know if you can desugar operators, like in many functional languages. E.g in OCaml you can do:
let x = (+) 3 5
The code above inits the variable x
with the value 3 + 5
. Writing (+)
is equivalent of having a local function that takes two parameters and returns their sum. (+) 3 5
is calling this function with the two arguments 3
and 5
.
The motivation behinds this is that you can pass the operator directly to a function without having to wrapped it in a function before:
local t = {"ab", "d", "c" }
local function op_greaterthan (a,b) return a>b end
table.sort (t, op_greaterthan) --would like to write: table.sort (t, (>))
Thanks!
Solution
You can't.
The Lua interpreter is very small, and it "makes shortcuts" when dealing with operators; for the parser, they are simply not "functions".
If you try to use an operator without its params, like this:
f(+)
Then the interpreter will throw a syntax error.
Due to this implementation, you are limited to the options already discussed: either you use a wrapper function (such as add
) or you pass in a string and so some kind of eval
, as in jpjacobs' solution.
OTHER TIPS
Yes you can (I don't see the point of it, and it will be slower, but it is possible):
do
local mem={}
function unsugar(op,a,b)
if mem[op] then
print('Fetched operation from memory')
return mem[op](a,b)
else
local f=loadstring('local a,b=...; return a '..op..' b')
mem[op]=f
return f(a,b)
end
end
end
print(unsugar('+',1,2)) -- = 3
print(unsugar('%',5,3)) -- = 2
print(unsugar('%',5,3)) -- = Fetched operation from memory \n 2
Edit: Eliminated stray globals a and b, and put in memoizing to improve performance, by compiling each operation only once.