In Lua, how do you find out the key an object is stored in?

Question

How would you print() out or find out the index of an object?

For example, if I spawned 20 random rock objects on screen into an array RockTable = {};<

Answer 1

There's another way you can do it, using metamethods. [Edited to allow you to remove values too]

t = {} -- Create your table, can be called anything
t.r_index = {} -- Holds the number value, i.e. t[1] = 'Foo'
t.r_table = {} -- Holds the string value, i.e. t['Foo'] = 1

mt = {} -- Create the metatable
mt.__newindex = function (self, key, value) -- For creating the new indexes
    if value == nil then -- If you're trying to delete an entry then
        if tonumber(key) then -- Check if you are giving a numerical index
            local i_value = self.r_index[key] -- get the corrosponding string index
            self.r_index[key] = nil -- Delete
            self.r_table[i_value] = nil
        else -- Otherwise do the same as above, but for a given string index
            local t_value = self.r_table[key]
            self.r_index[t_value] = nil
            self.r_table[key] = nil
        end
    else
        table.insert(self.r_index, tonumber(key), value) -- For t[1] = 'Foo'
        self.r_table[value] = key -- For t['Foo'] = 1
    end
end
mt.__index = function (self, key) -- Gives you the values back when you index them
    if tonumber(key) then
        return (self.r_index[key]) -- For For t[1] = 'Foo'
    else
        return (self.r_table[key]) -- For t['Foo'] = 1
    end
end

setmetatable(t, mt) -- Creates the metatable

t[1] = "Rock1" -- Set the values
t[2] = "Rock2"

print(t[1], t[2]) -- And *should* proove that it works
print(t['Rock1'], t['Rock2'])

t[1] = nil
print(t[1], t[2]) -- And *should* proove that it works
print(t['Rock1'], t['Rock2'])

It's more versatile as you can copy the t value and take it with you; it also means that you only have to play around with the one variable most of the time - hopefully should reduce the likelihood of you trying to access the wrong thing.