[2.0.12] Copper cable inexplicably missing from arithmetic combinator output

Things that has been reported already before.
Mm2PL
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[2.0.12] Copper cable inexplicably missing from arithmetic combinator output

Post by Mm2PL »

The set up
Image of the setup.
Image of the setup.
2024-10-30_23-41.png (207.33 KiB) Viewed 249 times
Blueprint of this setup:

Expected result
I would expect both signals being seen in the output.

In my imagination this would perform (although probably in an optimised form):

Code: Select all

// assume out, in_red, in_green are maps of signal identifier to value
for(auto signal : ALL_SIGNALS) {
  int32_t result = in_red[signal] & in_green[signal];
  out[signal] = result;
}
// in this case this should be equivalent to
out[COPPER_CABLE] = 66'000 & 0xFFFF_FFFF;
out[HOLMIUM_PLATE] = 66'000 & 0xFFFF_FFFF;
This should result in both values being the same.
Actual result
The Copper Cable either results in 0 or isn't set, which is weird and unexpected. I've only managed to replicate this behavior with the Copper Cable. It just works TM with other signals I've tried.
Muche
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Re: [2.0.12] Copper cable inexplicably missing from arithmetic combinator output

Post by Muche »

This is related to Unable to assign Copper Wire recipe in Assembly Machine 2.

66k in the bottom combinator is set as signal Copper cable, whereas -1 in the top one is set as signal Copper wire. Unifying them makes the arithmetic combinator output the correct expected result of two signals.
Mm2PL
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Re: [2.0.12] Copper cable inexplicably missing from arithmetic combinator output

Post by Mm2PL »

Muche wrote: Wed Oct 30, 2024 11:50 pm 66k in the bottom combinator is set as signal Copper cable, whereas -1 in the top one is set as signal Copper wire. Unifying them makes the arithmetic combinator output the correct expected result of two signals.
Funny, that I never noticed that. Thank you for solving my problem. I would be highly in favor of changing the textures and names of these items to be different. They are too easy to confuse especially that they were one and the same previously.
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