4 ms·

Can someone who lives in the world of C fill in the blanks? If this was a major typo, I have done things along those lines.

justinweiss 16y ago
memset sets a block of memory to a given value. It is often used to initialize an uninitialized block of memory to 0. So often, in fact, that this bug where it always initialized the memory to 0 survived so long.
- mey 16y ago
  Wish there was a cross-link to a bug tracking system where this was initially reported.
  - pufuwozu 16y ago
    You can just search for the change ID: https://review.source.android.com/#change,14699 https://review.source.android.com/#change,14699
- abstractbill 16y ago
  So often, in fact, that this bug where it always initialized the memory to 0 survived so long. Given that the bug went undetected for so long, I think you could almost reasonably claim that calling memset with a nonzero parameter is a "corner case" ;)
  - Psyonic 16y ago
    It is a "corner case," or close to it, but that's not why it wasn't caught here. I don't think this code is ever called, as explained above. It's default code in case there isn't a platform specific version, but there pretty much always is a platform specific version. Link: http://news.ycombinator.com/item?id=1379639 http://news.ycombinator.com/item?id=1379639
- fserb 16y ago
  on Google code search: "lang:c memset\(.+,\s[^0]+\s,.+\)" return 32k results and "lang:c memset\(.+,\s0+\s,.+\)" returns 335k.
  - wlievens 16y ago
    10% is a lot more than I would have thought, wow. Good job looking that up.
amock 16y ago
The function was always setting the memory to 0 instead of what was passed in. I think most uses of memset are to initialize memory to 0, but any other use would have had unexpected results.
draebek 16y ago
memset's job is to fill a block of memory with a given byte. The most common fill byte is 0, to initialize or otherwise clear a buffer. It looks like it is so common, in fact, that a programmer accidentally made Android's memset _always_ fill with 0, no matter what byte was passed in, and presumably no one noticed for a while. (I can't figure out when this patch was introduced.)
- jared314 16y ago
  The "Up to change" link takes you to the summary. May 13, 2010
fourneau 16y ago
Memset (according to the c++ reference) is supposed to: "Sets the first num bytes of the block of memory pointed by ptr to the specified value (interpreted as an unsigned char)." So, give it a pointer to some memory, give it a value and the number of bytes you want to set and it'll go wild. In this case, memset was just blindly dropping in zeroes into the memory block

justinweiss 16y ago

memset sets a block of memory to a given value. It is often used to initialize an uninitialized block of memory to 0. So often, in fact, that this bug where it always initialized the memory to 0 survived so long.

mey 16y ago
Wish there was a cross-link to a bug tracking system where this was initially reported.
- pufuwozu 16y ago
  You can just search for the change ID: https://review.source.android.com/#change,14699 https://review.source.android.com/#change,14699
abstractbill 16y ago
So often, in fact, that this bug where it always initialized the memory to 0 survived so long. Given that the bug went undetected for so long, I think you could almost reasonably claim that calling memset with a nonzero parameter is a "corner case" ;)
- Psyonic 16y ago
  It is a "corner case," or close to it, but that's not why it wasn't caught here. I don't think this code is ever called, as explained above. It's default code in case there isn't a platform specific version, but there pretty much always is a platform specific version. Link: http://news.ycombinator.com/item?id=1379639 http://news.ycombinator.com/item?id=1379639
fserb 16y ago
on Google code search: "lang:c memset\(.+,\s[^0]+\s,.+\)" return 32k results and "lang:c memset\(.+,\s0+\s,.+\)" returns 335k.
- wlievens 16y ago
  10% is a lot more than I would have thought, wow. Good job looking that up.

amock 16y ago

The function was always setting the memory to 0 instead of what was passed in. I think most uses of memset are to initialize memory to 0, but any other use would have had unexpected results.

draebek 16y ago

memset's job is to fill a block of memory with a given byte. The most common fill byte is 0, to initialize or otherwise clear a buffer. It looks like it is so common, in fact, that a programmer accidentally made Android's memset _always_ fill with 0, no matter what byte was passed in, and presumably no one noticed for a while. (I can't figure out when this patch was introduced.)

jared314 16y ago
The "Up to change" link takes you to the summary. May 13, 2010

fourneau 16y ago

Memset (according to the c++ reference) is supposed to: "Sets the first num bytes of the block of memory pointed by ptr to the specified value (interpreted as an unsigned char)." So, give it a pointer to some memory, give it a value and the number of bytes you want to set and it'll go wild. In this case, memset was just blindly dropping in zeroes into the memory block