在我的大学，我们有一个关于某些内核编程的课程。我偶然发现由故障模块生成的oops消息，我想知道如何调用堆栈。
当我执行一些步骤来获得分段错误时，我有类似的东西：
469.541182] kernel BUG at mm/slub.c:2969!
[ 469.541187] invalid opcode: 0000 [#1] SMP
[ 469.541193] last sysfs file: /sys/devices/LNXSYSTM:00/LNXSYBUS:00/ACPI0003:00/power_supply/ACAD/online
[ 469.541198] Modules linked in: broken_module binfmt_misc vsock vmmemctl acpiphp snd_ens1371 gameport snd_ac97_codec ac97_bus snd_pcm_oss snd_mixer_oss snd_pcm snd_seq_dummy snd_seq_oss snd_seq_midi snd_rawmidi snd_seq_midi_event snd_seq snd_timer snd_seq_device ppdev snd vmci fbcon tileblit lp parport_pc intel_agp soundcore psmouse font bitblit softcursor serio_raw parport snd_page_alloc i2c_piix4 agpgart vga16fb vgastate shpchp pcnet32 mptspi mptscsih floppy mii mptbase scsi_transport_spi vmxnet vmw_pvscsi vmxnet3 [last unloaded: broken_module]
[ 469.541259]
[ 469.541267] Pid: 2463, comm: cat Tainted: G W (2.6.32.46+drm33.20 #3) VMware Virtual Platform
[ 469.541274] EIP: 0060:[<c01fa7a5>] EFLAGS: 00210246 CPU: 0
[ 469.541286] EIP is at kfree+0xf5/0x100
[ 469.541290] EAX: 00000000 EBX: c1912360 ECX: 00000000 EDX: 0889b000
[ 469.541295] ESI: 0889b000 EDI: f47fdf98 EBP: f47fdf3c ESP: f47fdf24
[ 469.541299] DS: 007b ES: 007b FS: 00d8 GS: 00e0 SS: 0068
[ 469.541305] Process cat (pid: 2463, ti=f47fc000 task=f69af2c0 task.ti=f47fc000)
[ 469.541308] Stack:
[ 469.541310] 00000060 00210206 ecc27e00 00000000 0889b000 f47fdf98 f47fdf64 f817b1d9
[ 469.541320] <0> 00000000 f47fdf5c bfe4ff3c 00008000 00000023 ecc27200 00008000 0889b000
[ 469.541330] <0> f47fdf8c c020609f f47fdf98 f47fdfb4 0804ce10 f817b170 c0583760 ecc27200
[ 469.541340] Call Trace:
[ 469.541351] [<f817b1d9>] ? broken_read+0x69/0xf0 [broken_module]
[ 469.541360] [<c020609f>] ? vfs_read+0x9f/0x1a0
[ 469.541366] [<f817b170>] ? broken_read+0x0/0xf0 [broken_module]
[ 469.541375] [<c0583760>] ? do_page_fault+0x160/0x3a0
[ 469.541383] [<c0206252>] ? sys_read+0x42/0x70
[ 469.541391] [<c010340c>] ? syscall_call+0x7/0xb
[ 469.541397] [<c0580000>] ? mutex_lock_interruptible+0x0/0x40
[ 469.541400] Code: 8b 03 e9 78 ff ff ff 8b 3d 30 a2 78 c0 85 ff 0f 84 48 ff ff ff 8b 0f 83 c7 04 89 f2 89 d8 ff d1 8b 0f 85 c9 75 f1 e9 32 ff ff ff <0f> 0b eb fe 8d b4 26 00 00 00 00 55 89 e5 0f 1f 44 00 00 83 e8
[ 469.541510] EIP: [<c01fa7a5>] kfree+0xf5/0x100 SS:ESP 0068:f47fdf24
[ 469.541531] ---[ end trace 6acfcaf41ffa7314 ]---
我发现在cat进程中遇到了错误（这是有道理的，因为我执行的最后一步是cat /dev/broken）并且kfree函数出现了问题。
我在阅读跟踪时遇到了一些问题。
首先，我为什么要将broken_read功能列出两次？如果我查看源代码，则不会从broken_read到broken_read进行递归调用。我也想知道如何阅读0x69/0xf0。据我所知，第一个值是从函数开始的偏移量（在本例中为broken_read），但我不知道如何解释第二个值（0xf0）。
现在，当我执行objdump -S -d -r broken_module.ko时，我可以看到
00000170 <broken_read>:
int broken_release(struct inode *inode, struct file *filp) {
return 0;
}
ssize_t broken_read(struct file *filp, char *user_buf, size_t count, loff_t *f_pos) {
170: 55 push %ebp
171: 89 e5 mov %esp,%ebp
173: 83 ec 20 sub $0x20,%esp
176: 89 5d f4 mov %ebx,-0xc(%ebp)
179: 89 75 f8 mov %esi,-0x8(%ebp)
17c: 89 7d fc mov %edi,-0x4(%ebp)
17f: e8 fc ff ff ff call 180 <broken_read+0x10>
180: R_386_PC32 mcount
extern void *kmem_cache_alloc_notrace(struct kmem_cache *s, gfp_t gfpflags);
#else
static __always_inline void *
kmem_cache_alloc_notrace(struct kmem_cache *s, gfp_t gfpflags)
{
return kmem_cache_alloc(s, gfpflags);
184: b8 7c 04 00 00 mov $0x47c,%eax
185: R_386_32 kmalloc_caches
189: 8b 7d 08 mov 0x8(%ebp),%edi
18c: 89 d6 mov %edx,%esi
18e: ba d0 00 00 00 mov $0xd0,%edx
193: e8 fc ff ff ff call 194 <broken_read+0x24>
194: R_386_PC32 kmem_cache_alloc
198: 89 c3 mov %eax,%ebx
if (!s)
return ZERO_SIZE_PTR;
ret = kmem_cache_alloc_notrace(s, flags);
trace_kmalloc(_THIS_IP_, ret, size, s->size, flags);
19a: a1 80 04 00 00 mov 0x480,%eax
19b: R_386_32 kmalloc_caches
19f: 89 45 f0 mov %eax,-0x10(%ebp)
{(unsigned long)__GFP_THISNODE, "GFP_THISNODE"}, \
{(unsigned long)__GFP_RECLAIMABLE, "GFP_RECLAIMABLE"}, \
{(unsigned long)__GFP_MOVABLE, "GFP_MOVABLE"} \
) : "GFP_NOWAIT"
TRACE_EVENT(kmalloc,
1a2: a1 04 00 00 00 mov 0x4,%eax
1a3: R_386_32 __tracepoint_kmalloc
1a7: 85 c0 test %eax,%eax
1a9: 75 70 jne 21b <broken_read+0xab>
char* mybuf = NULL;
int mybuf_size = 100;
int len, err;
mybuf = kmalloc(mybuf_size, GFP_KERNEL);
if (!mybuf) {
1ab: 85 db test %ebx,%ebx
1ad: 74 60 je 20f <broken_read+0x9f>
broken_exit();
return -ENOMEM;
} else {
fill_buffer(mybuf,mybuf_size);
1af: ba 64 00 00 00 mov $0x64,%edx
1b4: 89 d8 mov %ebx,%eax
1b6: e8 fc ff ff ff call 1b7 <broken_read+0x47>
1b7: R_386_PC32 fill_buffer
}
len = strlen(mybuf);
1bb: 89 d8 mov %ebx,%eax
1bd: e8 fc ff ff ff call 1be <broken_read+0x4e>
1be: R_386_PC32 strlen
err = copy_to_user(user_buf, mybuf, len);
1c2: 89 da mov %ebx,%edx
1c4: 89 c1 mov %eax,%ecx
return -ENOMEM;
} else {
fill_buffer(mybuf,mybuf_size);
}
len = strlen(mybuf);
1c6: 89 45 f0 mov %eax,-0x10(%ebp)
err = copy_to_user(user_buf, mybuf, len);
1c9: 89 f0 mov %esi,%eax
1cb: e8 fc ff ff ff call 1cc <broken_read+0x5c>
1cc: R_386_PC32 copy_to_user
1d0: 89 c3 mov %eax,%ebx
kfree(user_buf);
1d2: 89 f0 mov %esi,%eax
1d4: e8 fc ff ff ff call 1d5 <broken_read+0x65>
1d5: R_386_PC32 kfree
read_count++;
1d9: 83 05 00 00 00 00 01 addl $0x1,0x0
1db: R_386_32 read_count
if (!err && *f_pos == 0) {
1e0: 85 db test %ebx,%ebx
1e2: 75 1c jne 200 <broken_read+0x90>
1e4: 8b 47 04 mov 0x4(%edi),%eax
1e7: 0b 07 or (%edi),%eax
1e9: 75 15 jne 200 <broken_read+0x90>
return -ENOMEM;
} else {
fill_buffer(mybuf,mybuf_size);
}
len = strlen(mybuf);
1eb: 8b 45 f0 mov -0x10(%ebp),%eax
read_count++;
if (!err && *f_pos == 0) {
*f_pos += len;
1ee: 89 c2 mov %eax,%edx
1f0: c1 fa 1f sar $0x1f,%edx
1f3: 89 07 mov %eax,(%edi)
1f5: 89 57 04 mov %edx,0x4(%edi)
return len;
1f8: eb 08 jmp 202 <broken_read+0x92>
1fa: 8d b6 00 00 00 00 lea 0x0(%esi),%esi
200: 31 c0 xor %eax,%eax
} else {
return 0;
}
}
202: 8b 5d f4 mov -0xc(%ebp),%ebx
205: 8b 75 f8 mov -0x8(%ebp),%esi
208: 8b 7d fc mov -0x4(%ebp),%edi
20b: 89 ec mov %ebp,%esp
20d: 5d pop %ebp
20e: c3 ret
int mybuf_size = 100;
int len, err;
mybuf = kmalloc(mybuf_size, GFP_KERNEL);
if (!mybuf) {
broken_exit();
20f: e8 fc ff ff ff call 210 <broken_read+0xa0>
210: R_386_PC32 broken_exit
214: b8 f4 ff ff ff mov $0xfffffff4,%eax
return -ENOMEM;
219: eb e7 jmp 202 <broken_read+0x92>
21b: 8b 15 10 00 00 00 mov 0x10,%edx
21d: R_386_32 __tracepoint_kmalloc
221: 85 d2 test %edx,%edx
223: 89 55 ec mov %edx,-0x14(%ebp)
226: 74 83 je 1ab <broken_read+0x3b>
228: 8b 02 mov (%edx),%eax
22a: 89 45 e8 mov %eax,-0x18(%ebp)
22d: 8b 55 f0 mov -0x10(%ebp),%edx
230: b8 70 01 00 00 mov $0x170,%eax
231: R_386_32 .text
235: b9 64 00 00 00 mov $0x64,%ecx
23a: c7 44 24 04 d0 00 00 movl $0xd0,0x4(%esp)
241: 00
242: 89 14 24 mov %edx,(%esp)
245: 89 da mov %ebx,%edx
247: ff 55 e8 call *-0x18(%ebp)
24a: 83 45 ec 04 addl $0x4,-0x14(%ebp)
24e: 8b 45 ec mov -0x14(%ebp),%eax
251: 8b 00 mov (%eax),%eax
253: 85 c0 test %eax,%eax
255: 89 45 e8 mov %eax,-0x18(%ebp)
258: 75 d3 jne 22d <broken_read+0xbd>
25a: e9 4c ff ff ff jmp 1ab <broken_read+0x3b>
25f: 90 nop
（我用EXTRA_CFLAGS+=g编译了模块）
现在，如果我将0x69添加到0x170，我会收到1D9，并且在列表中我看到了
1d0: 89 c3 mov %eax,%ebx
kfree(user_buf);
1d2: 89 f0 mov %esi,%eax
1d4: e8 fc ff ff ff call 1d5 <broken_read+0x65>
1d5: R_386_PC32 kfree
read_count++;
1d9: 83 05 00 00 00 00 01 addl $0x1,0x0
1db: R_386_32 read_count
所以在1d9下有变量递增。所以我想知道1d9是否意味着“控制返回此函数时执行指令的地址”？