Lua variable variables

Question

I'm very new to Lua, I'm working with it on an io controller which has strict limits on script sizes so I need to work within these limits.

I have a number of relays which I am controlling (1-64). I need to switch off a relay when an event happens, but the relay I'm switching off can change. I have a variable that holds the relay number and I need to turn off this relay.

I can achieve this using I statements:

if variable = 1 then
  io.relay1=0 //Turns off the relay
end
else if variable = 2 then
  io.relay2=0 //Turns off the relay
end

However, this will very quickly become a large script when repeated for the 64 relays. Is it possible to address the relay using the value of the variable as the relay name? Similar to the below:

io.relay{variable}=0 //Turns off the relay1/2/3/4/5 etc. depending on the value of variable

Alternatively, is there another way to keep the code compact?

lhf · Accepted Answer · 2018-04-20 16:08:21Z

3

Use

io["relay".. variable]=0

However, this creates a string every time.

If you can change how io works, a better solution would be to make io.relay a table and then simply do io.relay[variable]=0.

answered Apr 20, 2018 at 16:08

lhf

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2 Comments

Mike V. Over a year ago

relays["relay"..tostring(variable)] = 0 - if variable is a nil

James Holland Over a year ago

Absolutely spot on, thank you for your help. The io module is off the shelf, I have no control over the naming or structure of the addressing. Your suggestion works perfectly. Thank you

Paul Belanger · Accepted Answer · 2018-04-23 12:46:41Z

0

To avoid the string allocation issue that lhf's answer has, you could pre-generate a string table and index into that:

relay_names = {}
for k = 1,64 do
  relay_names[k] = "relay"..tostring(k)
end

Then setting the IO states would look something like this:

io[relay_names[1]] = 1

answered Apr 23, 2018 at 12:46

Paul Belanger

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