日期:2014-05-16 浏览次数:20806 次
试定位一个coredump的例子来验证一下。
堆栈:
(gdb) bt #0 0x43756109 in __memset_sse2 () from /lib/libc.so.6 #1 0x08048643 in main ()
汇编:
(gdb) frame 1 #1 0x08048643 in main () (gdb) disassemble Dump of assembler code for function main: 0x080485c0 <+0>: push %ebp 0x080485c1 <+1>: mov %esp,%ebp 0x080485c3 <+3>: and $0xfffffff0,%esp 0x080485c6 <+6>: sub $0x30,%esp 0x080485c9 <+9>: movl $0x0,0x18(%esp) 0x080485d1 <+17>: movl $0x0,0x1c(%esp) 0x080485d9 <+25>: movl $0x0,0x20(%esp) 0x080485e1 <+33>: movl $0x0,0x24(%esp) 0x080485e9 <+41>: movl $0x0,0x2c(%esp) 0x080485f1 <+49>: jmp 0x804860e <main+78> 0x080485f3 <+51>: movl $0x20,(%esp) 0x080485fa <+58>: call 0x8048490 <_Znaj@plt> 0x080485ff <+63>: mov %eax,%edx 0x08048601 <+65>: mov 0x2c(%esp),%eax 0x08048605 <+69>: mov %edx,0x18(%esp,%eax,4) 0x08048609 <+73>: addl $0x2,0x2c(%esp) 0x0804860e <+78>: cmpl $0x3,0x2c(%esp) 0x08048613 <+83>: setle %al 0x08048616 <+86>: test %al,%al 0x08048618 <+88>: jne 0x80485f3 <main+51> 0x0804861a <+90>: movb $0x0,0x2b(%esp) 0x0804861f <+95>: jmp 0x8048648 <main+136> 0x08048621 <+97>: movsbl 0x2b(%esp),%edx 0x08048626 <+102>: movsbl 0x2b(%esp),%eax 0x0804862b <+107>: mov 0x18(%esp,%eax,4),%eax 0x0804862f <+111>: movl $0x20,0x8(%esp) 0x08048637 <+119>: mov %edx,0x4(%esp) 0x0804863b <+123>: mov %eax,(%esp) 0x0804863e <+126>: call 0x8048470 <memset@plt> => 0x08048643 <+131>: addb $0x1,0x2b(%esp) 0x08048648 <+136>: cmpb $0x3,0x2b(%esp) 0x0804864d <+141>: setle %al 0x08048650 <+144>: test %al,%al 0x08048652 <+146>: jne 0x8048621 <main+97> 0x08048654 <+148>: mov $0x0,%eax 0x08048659 <+153>: jmp 0x8048663 <main+163> 0x0804865b <+155>: mov %eax,(%esp) 0x0804865e <+158>: call 0x80484b0 <_Unwind_Resume@plt> 0x08048663 <+163>: leave 0x08048664 <+164>: ret End of assembler dump.
由于coredump是在这一条指令下出错:
0x08048643 <+131>: addb $0x1,0x2b(%esp)
由memset的原型:
void *memset(void *s, int c, size_t n);
可知,会出现问题,要么,是第一个参数s非法,要么是n超出s的范围。
先看一下s是哪个,由
0x08048626 <+102>: movsbl 0x2b(%esp),%eax 0x0804862b <+107>: mov 0x18(%esp,%eax,4),%eax
和
0x0804863b <+123>: mov %eax,(%esp)
可知,
s的值存放在esp+0x18+eax*4。而eax的值是由esp+0x2b得来的。
由movsbl可知,esp+0x2b存放着一个char型,所以,
(gdb) x /c $esp+0x2b 0xbf88c15b: 1 '\001'
由
0x08048621 <+97>: movsbl 0x2b(%esp),%edx 0x08048626 <+102>: movsbl 0x2b(%esp),%eax 0x0804862b <+107>: mov 0x18(%esp,%eax,4),%eax 0x0804862f <+111>: movl $0x20,0x8(%esp) 0x08048637 <+119>: mov %edx,0x4(%esp) 0x0804863b <+123>: mov %eax,(%esp) 0x0804863e <+126>: call 0x8048470 <memset@plt> => 0x08048643 <+131>: addb $0x1,0x2b(%esp) 0x08048648 <+136>: cmpb $0x3,0x2b(%esp) 0x0804864d <+141>: setle %al 0x08048650 <+144>: test %al,%al 0x08048652 <+146>: jne 0x8048621 <main+97>
这个循环可知,esp+0x2