论坛徽章:: 0

电梯直达

1楼 [收藏(0)] [报告]

发表于 2015-11-02 17:52 |只看该作者 |倒序浏览

今天加载一个enc内核模块，碰到一个内存分配失败的问题，想请教一下，

出错的时，func_a调用了kzalloc（1M，GFP_KERNEL），根据之前的知识，我的理解是这是slab分配器分配失败的原因，但1M内存对应于order 8，而
“Normal: 1203*4kB 169*8kB 85*16kB 48*32kB 19*64kB 1*128kB 1*256kB 0*512kB 1*1024kB 0*2048kB 0*4096kB = 11684kB”显示buddy system有一块大小为1M的连续内存。我的疑问是，为什么slab分配器不能够从buddy中分配这块内存呢？

OOM打印如下：

insmod: page allocation failure: order:8, mode:0xd0
[ 19.860000] Backtrace:
[ 19.860000] [<8001126c>] (dump_backtrace+0x0/0x10c) from [<8037c248>] (dump_stack+0x18/0x1c)
[ 19.880000] r6:00000000 r5:000000d0 r4:00000001 r3:8048fc38
[ 19.880000] [<8037c230>] (dump_stack+0x0/0x1c) from [<8006d6d4>] (warn_alloc_failed+0xf4/0x11

[ 19.900000] [<8006d5e0>] (warn_alloc_failed+0x0/0x11

from [<8006f8cc>] (__alloc_pages_nodemask+0x580/0x5dc)
[ 19.920000]  r3:81bddd2c r2:00000000
[ 19.920000]  r7:00000008 r6:00000001 r5:81bdc000 r4:000000d0
[ 19.940000] [<8006f34c>] (__alloc_pages_nodemask+0x0/0x5dc) from [<8037d2bc>] (cache_alloc_refill+0x2d8/0x5d0)
[ 19.960000] [<8037cfe4>] (cache_alloc_refill+0x0/0x5d0) from [<800964c8>] (kmem_cache_alloc+0x64/0xb4)
[ 19.980000] [<80096464>] (kmem_cache_alloc+0x0/0xb4) from [<7f02c024>] (func_a+0x158/0x2b8 [enc])
[ 20.000000]  r7:7f03de40 r6:01800000 r5:80496ac4 r4:7f03e194
[ 20.000000] [<7f02becc>] (func_a+0x0/0x2b8 [enc]) from [<7f02e168>] (func_b+0x54/0x2b0 [enc])
[ 20.020000]  r7:7f03de40 r6:00000000 r5:7f045094 r4:804afe58
[ 20.020000] [<7f02e114>] (func_b+0x0/0x2b0 [enc]) from [<7f0450b4>] (init_func+0x20/0xbc [enc])
[ 20.040000]  r7:81b88420 r6:00000000 r5:7f045094 r4:7f03d0e8
[ 20.040000] [<7f045094>] (init_func+0x0/0xbc [enc]) from [<80008564>] (do_one_initcall+0x98/0x16

[ 20.060000] r5:7f045094 r4:7f03dcc8
[ 20.060000] [<800084cc>] (do_one_initcall+0x0/0x16

from [<8005840c>] (sys_init_module+0x1580/0x176c)
[ 20.080000]  r9:00000024 r8:00000001 r7:81b88420 r6:7f03dd10 r5:00000001
[ 20.080000] r4:7f03dcc8
[ 20.100000] [<80056e8c>] (sys_init_module+0x0/0x176c) from [<8000dac0>] (ret_fast_syscall+0x0/0x30)
[ 20.120000] Mem-info:
[ 20.120000] Normal per-cpu:
[ 20.130000] CPU 0: hi: 6, btch: 1 usd: 4
[ 20.140000] active_anon:136 inactive_anon:2541 isolated_anon:0
[ 20.140000]  active_file:139 inactive_file:148 isolated_file:0
[ 20.140000]  unevictable:0 dirty:0 writeback:0 unstable:0
[ 20.140000]  free:2914 slab_reclaimable:147 slab_unreclaimable:803
[ 20.140000]  mapped:104 shmem:2544 pagetables:32 bounce:0
[ 20.150000] Normal free:11656kB min:732kB low:912kB high:1096kB active_anon:544kB inactive_anon:10164kB active_file:556kB inactive_file:592kB unevictable:0kB isolated(anon):0kB isolated(file):0kB present:33528kB mlocked:0kB dirty:0kB writeback:0kB mapped:416kB shmem:10180kB slab_reclaimable:588kB slab_unreclaimable:3212kB kernel_stack:296kB pagetables:128kB unstable:0kB bounce:0kB writeback_tmp:0kB pages_scanned:7 all_unreclaimable? no
[ 20.170000] lowmem_reserve[]: 0 0
[ 20.170000] Normal: 1203*4kB 169*8kB 85*16kB 48*32kB 19*64kB 1*128kB 1*256kB 0*512kB 1*1024kB 0*2048kB 0*4096kB = 11684kB
[ 20.180000] 2825 total pagecache pages
[ 20.190000] 0 pages in swap cache
[ 20.190000] Swap cache stats: add 0, delete 0, find 0/0
[ 20.200000] Free swap  = 0kB
[ 20.200000] Total swap = 0kB
[ 20.210000] 8448 pages of RAM
[ 20.210000] 2989 free pages
[ 20.220000] 1298 reserved pages
[ 20.220000] 951 slab pages
[ 20.220000] 544 pages shared
[ 20.230000] 0 pages swap cached
[ 20.230000] SLAB: Unable to allocate memory on node 0 (gfp=0xd0)
[ 20.240000] cache: size-1048576, object size: 1048576, order: 8
[ 20.250000] node 0: slabs: 1/1, objs: 1/1, free: 0
insmod: can't insert '/mnt/enc.ko': Cannot allocate memory

文库|博客

Godbach

版主

论坛徽章:: 36

2楼 [报告]

发表于 2015-11-02 18:45 |只看该作者

回复 1# alexhak2004

建议看一下 kzalloc 的实现，这个应该比较好跟。

据我以前老的内核知识来讲，内核中的连续大块内存，是比较珍贵的资源。你这 1M 属于比较大的了，如果没有特别需要，可以考虑用 vmalloc。

实战分享：从技术角度谈机器学习入门| 【大话IT】RadonDB低门槛向MySQL集群下战书 | ChinaUnix打赏功能已上线！ | 新一代分布式关系型数据库RadonDB知多少？

alexhak2004

白手起家

论坛徽章:: 0

3楼 [报告]

发表于 2015-11-03 09:35 |只看该作者

谢谢回复，这里之所以使用kzalloc是因为给硬件使用，需要连续，所以vmalloc不能满足要求。

而我的疑问是，从打印“Normal: 1203*4kB 169*8kB 85*16kB 48*32kB 19*64kB 1*128kB 1*256kB 0*512kB 1*1024kB 0*2048kB 0*4096kB”看，buddy中是有一块大小为1M的连续内存的，但为何分配不出来。