[Python-Dev] Wordcode: new regular bytecode using 16-bit units

[Victor Stinner]

2016-04-13 23:02 GMT+02:00 Eric Fahlgren <ericfahlgren at gmail.com>:
> Percentage of 1-byte args    = 96.80%

Yeah, I expected such high ratio. Good news that you confirm it.


> Non-argument ops             =    53,719
> One-byte args                =   368,787
> Multi-byte args              =    12,191

Again, only a very few arguments take multiple bytes. Good, the
bytecode will be smaller.

IMHO it's more a nice side effect than a real goal. The runtime
performance matters more than the size of the bytecode, it's not like
a bytecode take 4 MB. It's probably closer to 1 KB and so can probably
benefit of the fatest CPU caches.


> Just for the record, here's my arithmetic:
> byteCodeSize     = 1*nonArgumentOps + 3*oneByteArgs + 3*multiByteArgs
> wordCodeSize     = 2*nonArgumentOps + 2*oneByteArgs + 4*multiByteArgs

If multiByteArgs means any size > 1 byte, the wordCodeSize formula is wrong:

- no parameter: 2 bytes
- 8-bit parameter: 2 bytes
- 16-bit parameter: 4 bytes
- 24-bit parameter: 6 bytes
- 32-bit parameter: 8 bytes

But you wrote that you didn't see EXTEND_ARG, so I guess that
multibyte means 16-bit in your case, and so your formula is correct.

Hopefully, I don't expect 32-bit parameters in the wild, only 24-bit
parameter for function with annotation.


> (It is interesting to note that I have never encountered an EXTENDED_ARG operator in the wild, only in my own synthetic examples.)

As I wrote, EXTENDED_ARG can be seen when MAKE_FUNCTION is used with
annotations.

Victor