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Hmm, I actually have no idea.

Or you just calculate the block message for each possible message, to a XOR compare between message received and block message and count the number of ones that differ, each one is an error. Choose the possible message with least ones. This can certainly be made with a LUT (like the C++ code above). At least it doesn't require 64*313 billion if statements...

But a more elegant way... Hmm, I actually have no idea.

Subtract 32 from 40, and you get 8. For (1) it was the 8th bit (from left) that had an error. You can pinpoint which one that is wrong.

for (2) there's 2 bits of errors and you are always below 32. You can't.... really pinpoint which two that are wrong... There's not enough information from the parity multiplication to deduce those two...

The (4) above and (8) above was copied and compared to other block messages when I used the generator matrix for values of 0 to 63. There was none. So if there are 4 errors.. then the parity multiplication will say 0... It feels as if I'm inventing the wheel... and failing.

Hmm, I'm out on deep water and have only used LUT's so far. And I've spent too much time on this as someone who has had a horrible teacher in this subject and I don't regularly work with this on a daily basis. => I'm not an expert in this subject. Oh well.

The (4) above and (8) above was copied and compared to other block messages when I used the generator matrix for values of 0 to 63. There were none that matched. So (4) and (8) above is unique to the number 43. So if there are 4 errors.. then the parity multiplication will say 0... It feels as if I'm inventing the wheel... and failing.

Hmm, I'm out on deep water and have only used LUT's so far. And I've spent too much time on this as someone who has had a horrible teacher in this subject and I don't regularly work with this on a daily basis. => I'm not an expert in this subject. Oh well.

If I were to solve it with k=5 => block messages with 64 bits. Then I can't use a LUT with \$2^{64}\$ rows. That's just silly. In all honesty I'd probably use a switch statement that only covers 0 and 7 bits error. The first switch statement in this answer only covers 0 bits of error and hopes that you fixed the wrong bit if there is one before you insert it. You can easily just add more "case (1): 43, case (2): 43, case(3):43 and etc... for all... shapes of 7 bit errors...... No wait... this will give you a ridiculous amount of cases just for the number 43. The different combination of 7 bits of error in a 64 bit message is n choose k => 64 choose 7 => \${64 \choose 7}=313092960768\approx 313 \text{ billion}\$ different ways. That's.. a sick amount of cases just to solve 7 bits of error for the number 43... No... this is not the right way...

Hmm, I actually have no idea.