Drum rack and lights firmware examples #1

					
				@ -0,0 +145,4 @@

				                if not col_pin.value():

				                    pressed_keys.add(key)

				            row_pin.value(1)

				        return set(pressed_keys)

This is a (minor) memory ~~leak~~ allocation inefficiency as it will create a new set object on every loop iteration. It's fine to return pressed_keys directly.

This is a (minor) memory ~~leak~~ allocation inefficiency as it will create a new `set` object on every loop iteration. It's fine to return `pressed_keys` directly.

Yes, makes sense to remove it. ~~Out of curiousity, shouldn't python take care of cleaning up that memory once the references to the new set go out of scope?~~ just saw your correction :)

Also, not sure if it's overoptimizing things, but we could use a single integer instead of a set here. Each bit representing an on/off state of a btn. We'd have to deal with bitwise operations, but the memory footprint would likely be smaller. Maybe the amount of cycles too?

E.g.

# no btns pressed
pressed_keys = 0

# btn 3 is pressed
pressed_keys = pressed_keys | (1 << 3)

# btn 4 is not pressed
pressed_keys = pressed_keys & ~(1 << 4)

# check if btn 2 is pressed
(pressed_keys >> 2) & 1

# set intersection
pressed_keys_1 & pressed_keys_2

# set difference
pressed_keys_1 & ~pressed_keys_2

# get all the pressed btns
pressed_list = []
for i in range(NUM_BTNS):
  if (pressed_keys >> i) & 1:
    pressed_list.append(i)

Yes, makes sense to remove it. ~Out of curiousity, shouldn't python take care of cleaning up that memory once the references to the new set go out of scope?~ just saw your correction :) Also, not sure if it's overoptimizing things, but we could use a single integer instead of a set here. Each bit representing an on/off state of a btn. We'd have to deal with bitwise operations, but the memory footprint would likely be smaller. Maybe the amount of cycles too? E.g. ```python # no btns pressed pressed_keys = 0 # btn 3 is pressed pressed_keys = pressed_keys | (1 << 3) # btn 4 is not pressed pressed_keys = pressed_keys & ~(1 << 4) # check if btn 2 is pressed (pressed_keys >> 2) & 1 # set intersection pressed_keys_1 & pressed_keys_2 # set difference pressed_keys_1 & ~pressed_keys_2 # get all the pressed btns pressed_list = [] for i in range(NUM_BTNS): if (pressed_keys >> i) & 1: pressed_list.append(i) ```

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viernullvier/neogrid!1

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