NodeMCU debounce based on timer with GPIO pullup

debounce-with-tmr.lua

-- inspired by https://github.com/hackhitchin/esp8266-co-uk/blob/master/tutorials/introduction-to-gpio-api.md
-- and http://www.esp8266.com/viewtopic.php?f=24&t=4833&start=5#p29127
local pin = 4    --> GPIO2

function debounce (func)
    local last = 0
    local delay = 50000 -- 50ms * 1000 as tmr.now() has μs resolution

    return function (...)
        local now = tmr.now()
        local delta = now - last
        if delta < 0 then delta = delta + 2147483647 end; -- proposed because of delta rolling over, https://github.com/hackhitchin/esp8266-co-uk/issues/2
        if delta < delay then return end;

        last = now
        return func(...)
    end
end

function onChange ()
    print('The pin value has changed to '..gpio.read(pin))
end

gpio.mode(pin, gpio.INT, gpio.PULLUP) -- see https://github.com/hackhitchin/esp8266-co-uk/pull/1
gpio.trig(pin, 'both', debounce(onChange))

I'm using the same routine for debouncing and the switch is having issues with controlling inductive loads. The button would get triggered as soon as the relay is switched remotely. I think the code would need to include some kind of logic to detect minimum time the button is pressed.

Also there is some chance (0.002%) that the code doesn't respond to button press due to the rollover of tmr.now() and the initial value of "last". Obviously not a problem for majority of applications but good to keep in mind.

@Bop4yN you're missing that debounce returns a function (line 9-17). So, every time the GPIO triggers it doesn't run debounce but the function it returned when it was once (i.e. initially) called. That inner function has access to last and keeps changing it.

-> https://en.wikipedia.org/wiki/Closure_(computer_programming)