Timeline for How to do a single pass with badblocks in destructive mode?
Current License: CC BY-SA 4.0
10 events
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Nov 7, 2022 at 15:35 | answer | added | lmaguina | timeline score: 0 | |
Aug 25, 2022 at 10:14 | history | edited | Jon Bentley | CC BY-SA 4.0 |
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Dec 6, 2019 at 21:02 | vote | accept | Jon Bentley | ||
May 5, 2015 at 12:17 | comment | added | Slizzered |
Actually you got it a bit wrong, I think. Writing itself can not be checked against failure. It writes 0xff and immediately checks if there is 0xff written by reading the location from disk. So the stuck 0 is ruled out. Then, it does the same with 0x00 to rule out the stuck 1 error. so 1 pass with each pattern is sufficient. The other ones (alternating pattern of 0 and 1 ) are against bits that stick together (if I set bit N to 1 , bit N+1 also goes to 1 as a result of the error)
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May 5, 2015 at 12:13 | comment | added | Jon Bentley | @Slizzered Thinking about it, I suppose that wouldn't cover the case where writing a 1 would fail, but the bit already has a 1, but I can't think of a scenario where 3 passes would be insufficient (0xff, 0x00, 0xff again) | |
May 5, 2015 at 12:09 | comment | added | Jon Bentley | @Slizzered Thanks, that's useful to consider. Shouldn't two passes be sufficient in that case? 0xff and 0x00. | |
May 5, 2015 at 11:41 | comment | added | Slizzered |
Keep in mind, that the multiple passes are not for erasing the data. They are necessary to find certain errors on your disk. Otherwise you might not find all bad blocks. Consider the case where a bit is always stuck at 1 . If you only did test with the 0xff -pattern, you would not be able to find this error. Since badblocks is not designed to erase the data, but to find bad blocks, multiple passes is not considered to be excessive.
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May 5, 2015 at 8:25 | history | edited | Jon Bentley | CC BY-SA 3.0 |
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May 5, 2015 at 7:43 | answer | added | Jon Bentley | timeline score: 11 | |
May 5, 2015 at 6:58 | history | asked | Jon Bentley | CC BY-SA 3.0 |