Lua : Dynamicly calling a function with arguments.

Posted by Tipx on Stack Overflow See other posts from Stack Overflow or by Tipx
Published on 2010-03-14T05:40:07Z Indexed on 2010/03/14 5:45 UTC
Read the original article Hit count: 318

Filed under:
|
|

Using Lua, I'm trying to dynamicly call a function with parameters. What I want to have it done is I send a string to be parsed in a way that :

  • 1st argument is a class instance "Handle"
  • 2nd is the function to be called
  • All that is left are arguments

"modules" is a a table like { string= } split() is a simple parser that returns a table with indexed strings

function Dynamic(msg)
    local args = split(msg, " ")
    module = args[1]
    table.remove(args, 1)
    if module then
        module = modules[module]
        command = args[1]
        table.remove(args, 1)
        if command then
            if not args then
                module[command]()
            else
                module[command](unpack(args))  -- Reference 1
            end
        else
            -- Function doesnt exist
        end
    else
        -- Module doesnt exist
    end
end

When I try this with "ignore remove bob", by "Reference 1", it tries to call "remove" on the instance associated with "ignore" in modules, and gives the argument "bob", contained in a table (with a single value).

However, on the other side of the call, the remove function does not receive the argument. I even tried to replace the "Reference 1" line with

module[command]("bob")

but I get the same result.

© Stack Overflow or respective owner

Related posts about lua

Related posts about dynamic