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In my workflow, I need to do binary operations quite often. I am wondering if there is a clean way of creating macro/elisp-function to return value of following HASH function.

  • Input 16bit hex number: NUM
  • Hash function return value = NUM[15] ^ NUM[13] ^ NUM[11] ^ ... ^ NUM[1] (here ^ means XOR)
  • For example:
    • If NUM(in hex)=0001, hash=0
    • if NUM(in hex)=8000, hash=1
    • if NUM(in hex)=000A, hash=0
  • This seems equivalent to ((NUM & 0x55) > 0)?1,0) in C syntax. i.e. check if any of the odd bits are 1. – Juancho Jun 7 '18 at 18:51
  • Hi Jiancho, not precisely. it's xor of all bits in odd position. Sorry, I'll edit the question to clarify. – StupidKris Jun 7 '18 at 18:55
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In calc, you can perform this operation as follows:

  • type your number to check (e.g. 16#32FE)
  • keep odd numbered bits: 16#AAAA b a. This ands your number with 0xAAAA
  • binary unpack with b u. This returns a list of active bits.
  • count length of list with V #
  • check for parity: 2%

Turning this into a macro... (currently reading the manual).

  • Thanks Juancho, this worked perfectly as I had expected. So many things yet to be explored in calc <3 – StupidKris Jun 8 '18 at 10:05

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