Christopher Haster
0cea8b96fb
Do you see the O(n^2) behavior in this loop?
j = 0
while j < len(data):
word, d = fromleb(data[j:])
j += d
The slice, data[j:], creates a O(n) copy every iteration of the loop.
A bit tricky. Or at least I found it tricky to notice. Maybe because
array indexing being cheap is baked into my brain...
Long story short, this repeated slicing resulted in O(n^2) behavior in
Rbyd.fetch and probably some other functions. Even though we don't care
_too_ much about performance in these scripts, having Rbyd.fetch run in
O(n^2) isn't great.
Tweaking all from* functions to take an optional index solves this, at
least on paper.
---
In practice I didn't actually find any measurable performance gain. I
guess array slicing in Python is optimized enough that the constant
factor takes over?
(Maybe it's being helped by us limiting Rbyd.fetch to block_size in most
scripts? I haven't tested NAND block sizes yet...)
Still, it's good to at least know this isn't a bottleneck.
90 lines
2.2 KiB
Python
Executable File
90 lines
2.2 KiB
Python
Executable File
#!/usr/bin/env python3
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# prevent local imports
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if __name__ == "__main__":
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__import__('sys').path.pop(0)
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import io
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import os
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import struct
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import sys
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# open with '-' for stdin/stdout
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def openio(path, mode='r', buffering=-1):
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import os
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if path == '-':
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if 'r' in mode:
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return os.fdopen(os.dup(sys.stdin.fileno()), mode, buffering)
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else:
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return os.fdopen(os.dup(sys.stdout.fileno()), mode, buffering)
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else:
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return open(path, mode, buffering)
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def fromle32(data, j=0):
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return struct.unpack('<I', data[j:j+4].ljust(4, b'\0'))[0]
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def dbg_le32s(data):
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lines = []
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j = 0
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while j < len(data):
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word = fromle32(data, j)
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lines.append((
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' '.join('%02x' % b for b in data[j:j+4]),
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word))
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j += 4
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# figure out widths
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w = [0]
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for l in lines:
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w[0] = max(w[0], len(l[0]))
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# then print results
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for l in lines:
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print('%-*s %s' % (
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w[0], l[0],
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l[1]))
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def main(le32s, *,
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hex=False,
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input=None):
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hex_ = hex; del hex
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# interpret as a sequence of hex bytes
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if hex_:
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bytes_ = [b for le32 in le32s for b in le32.split()]
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dbg_le32s(bytes(int(b, 16) for b in bytes_))
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# parse le32s in a file
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elif input:
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with openio(input, 'rb') as f:
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dbg_le32s(f.read())
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# we don't currently have a default interpretation
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else:
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print("error: no -x/--hex or -i/--input?",
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file=sys.stderr)
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sys.exit(1)
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if __name__ == "__main__":
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import argparse
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import sys
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parser = argparse.ArgumentParser(
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description="Decode le32s.",
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allow_abbrev=False)
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parser.add_argument(
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'le32s',
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nargs='*',
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help="Le32s to decode.")
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parser.add_argument(
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'-x', '--hex',
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action='store_true',
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help="Interpret as a sequence of hex bytes.")
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parser.add_argument(
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'-i', '--input',
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help="Read le32s from this file. Can use - for stdin.")
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sys.exit(main(**{k: v
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for k, v in vars(parser.parse_intermixed_args()).items()
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if v is not None}))
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