CODEWITHSUNDEEP
stringluaarguments
cevikarda
cevikarda

Reputation: 258

Lua argument as string name

I have a string like this:

local tempStr = "abcd"

and I want to send the variable which is named as "abcd" to a function like this:

local abcd = 3

print( tempStr ) -- not correct!!

and the result will be 3, not abcd.

Upvotes: 2

Views: 1217

Answers (3)

valker
valker

Reputation: 143

The function debug.getlocal could help you.

function f(name)
    local index = 1
    while true do
        local name_,value = debug.getlocal(2,index)
        if not name_ then break end
        if name_ == name then return value end
        index = index + 1
    end 
end

function g()
    local a = "a!"
    local b = "b!"
    local c = "c!"
    print (f("b")) -- will print "b!"
end

g()

Upvotes: 1

Nicol Bolas
Nicol Bolas

Reputation: 473202

You can't do that with variables declared as local. Such variables are just stack addresses; they don't have permanent storage.

What you're wanting to do is use the content of a variable to access an element of a table. Which can of course be the global table. To do that, you would do:

local tempStr = "abcd"
abcd = 3 --Sets a value in the global table.
print(_G[tempStr]) --Access the global table and print the value.

You cannot do that if you declare abcd as local.

Upvotes: 1

kikito
kikito

Reputation: 52621

You can do it with local variables if you use a table instead of a "plain" variable:

local tempStr = "abcd"

local t = {}

t[tempStr] = 3

print( t[tempStr]) -- will print 3

Upvotes: 3

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