[EMAIL PROTECTED] wrote:
I have written a program to do something similar. My strategy is:
* use os.read() to read 512 bytes at a time
* when os.read fails, seek to the next multiple of 512 bytes
and write '\0' * 512 to the output
I notice this code doesn't deal properly with short reads, but in my experience
they never happen (because the disk error makes an entire block unreadable, and
a block is never less than 512 bytes)
I use this code on a unix-style system.
def dd(src, target, bs=512):
src = os.open(src, os.O_RDONLY)
if os.path.exists(target):
target = os.open(target, os.O_WRONLY | os.O_APPEND, 0600)
existing = os.lseek(target, 0, SEEK_END)
else:
existing = 0
target = os.open(target, os.O_WRONLY | os.O_CREAT, 0600)
total = os.lseek(src, 0, SEEK_END) / bs
os.lseek(src, existing, SEEK_SET)
os.lseek(target, existing, SEEK_SET)
if existing: print "starting at", existing
i = existing / bs
f = 0
lastrem = -1
last = start = time.time()
while 1:
try:
block = os.read(src, bs)
except os.error, detail:
if detail.errno == errno.EIO:
block = "\0" * bs
os.lseek(src, (i+1) * bs, SEEK_SET)
f = f + 1
else:
raise
if block == "": break
i = i + 1
os.write(target, block)
now = time.time()
if i % 1000 or now - last < 1: continue
last = now
frac = i * 1. / total
rem = int((now-start) * (1-frac) / frac)
if rem < 60 or abs(rem - lastrem) > .5:
rm, rs = divmod(rem, 60)
lastrem = rem
spd = i * 512. / (now - start) / 1024 / 1024
sys.stderr.write("%8d %8d %8d %3.1f%% %6d:%02d %6.1fMB/s\r"
% (i, f, i-f, i * 100. / total, rm, rs, spd))
sys.stderr.write("\n")
Sorry but what are SEEK_END and SEEK_SET?
Maurice
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Doctor of Philosophy (Science) Candidate, The University of Melbourne
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