Assembly HW help, Assembly Language

Assignment Help:
I was wondering if you guys could offer me some advice and help on how to proceed - not answers- for a homework problem I am attempting. I am currently working on a "bomb" project in which I progress through stages by discovering passwords or phrases using the gdb debugger to analyze assembly code. I was able to easily complete stages 1-3, but am having a bit of trouble with stage 4. Here is the assembly for this stage:

0x08048d6d <+0>: sub $0x2c,%esp
0x08048d70 <+3>: lea 0x1c(%esp),%eax
0x08048d74 <+7>: mov %eax,0xc(%esp)
0x08048d78 <+11>: lea 0x18(%esp),%eax
0x08048d7c <+15>: mov %eax,0x8(%esp)
0x08048d80 <+19>: movl $0x804a5d1,0x4(%esp) ;%d, %d
0x08048d88 <+27>: mov 0x30(%esp),%eax
0x08048d8c <+31>: mov %eax,(%esp)
0x08048d8f <+34>: call 0x8048850 <__isoc99_sscanf@plt>
0x08048d94 <+39>: cmp $0x2,%eax
0x08048d97 <+42>: jne 0x8048da5
0x08048d99 <+44>: mov 0x1c(%esp),%eax
0x08048d9d <+48>: sub $0x2,%eax
0x08048da0 <+51>: cmp $0x2,%eax
0x08048da3 <+54>: jbe 0x8048daa
0x08048da5 <+56>: call 0x80492f5
0x08048daa <+61>: mov 0x1c(%esp),%eax
0x08048dae <+65>: mov %eax,0x4(%esp)
0x08048db2 <+69>: movl $0x8,(%esp)
0x08048db9 <+76>: call 0x8048d23
0x08048dbe <+81>: cmp 0x18(%esp),%eax
0x08048dc2 <+85>: je 0x8048dc9
0x08048dc4 <+87>: call 0x80492f5
0x08048dc9 <+92>: add $0x2c,%esp
0x08048dcc <+95>: ret
From what I have looked at so far, it looks like this phase accepts two decimals (line 19), the second one must be less than or equal to 4, but greater than 1 (lines 48, 51, 54). Another important aspect to this problem is the inclusion of a recursive "func4" which is as follows:

0x08048d23 <+0>: push %edi
0x08048d24 <+1>: push %esi
0x08048d25 <+2>: push %ebx
0x08048d26 <+3>: sub $0x10,%esp
0x08048d29 <+6>: mov 0x20(%esp),%ebx
0x08048d2d <+10>: mov 0x24(%esp),%esi
0x08048d31 <+14>: test %ebx,%ebx
0x08048d33 <+16>: jle 0x8048d61
0x08048d35 <+18>: mov %esi,%eax
0x08048d37 <+20>: cmp $0x1,%ebx
0x08048d3a <+23>: je 0x8048d66
0x08048d3c <+25>: mov %esi,0x4(%esp)
0x08048d40 <+29>: lea -0x1(%ebx),%eax
0x08048d43 <+32>: mov %eax,(%esp)
0x08048d46 <+35>: call 0x8048d23
0x08048d4b <+40>: lea (%eax,%esi,1),%edi
0x08048d4e <+43>: mov %esi,0x4(%esp)
0x08048d52 <+47>: sub $0x2,%ebx
0x08048d55 <+50>: mov %ebx,(%esp)
0x08048d58 <+53>: call 0x8048d23
0x08048d5d <+58>: add %edi,%eax
0x08048d5f <+60>: jmp 0x8048d66
0x08048d61 <+62>: mov $0x0,%eax
0x08048d66 <+67>: add $0x10,%esp
0x08048d69 <+70>: pop %ebx
0x08048d6a <+71>: pop %esi
0x08048d6b <+72>: pop %edi
0x08048d6c <+73>: ret
This function is where I am having the most trouble...I am totally clueless as to what it does. All I think I know is that it accepts my second input value as an argument, alters in some way and finally compares it to the value 0x18(%esp) (line 81 in the first section)

The project is not due until later this week, but I would really rather have a better understanding of this material asap as right now I am a little lost.Thank you guys for your time.

Related Discussions:- Assembly HW help

Icwi-microprocessor, The definitions of the bits in ICWI are following: ...

The definitions of the bits in ICWI are following: Always set to the value 1. It directs the received byte to ICWI as oppose to OCW2 or OCW3. Which also utilize the even addr

Assembly Language Program, which uses BIOS interrupt INT 21 to read current...

which uses BIOS interrupt INT 21 to read current system time and displays it on the top-left corner of screen.

Movsw/movsb-string manipulation instruction-microprocessor, MOVSW/MOVSB : ...

MOVSW/MOVSB : Move String Word or String Byte: Imagine a string of bytes, stored in a set  of consecutive memory locations is to be moved to another set of  the destination locati

The intel processors , The Intel Processors :         The Intel Co...

The Intel Processors :         The Intel Corporation is the biggest manufacturer  of microchips  in the world,  in addition  to being  the leading provider of chips for PCs. I

Zero flag, Zero flag: The next line compares the value in register. A ...

Zero flag: The next line compares the value in register. A with the value 1. If they are equivalent, the Zero flag is set (to 1). The next line then jumps to start: only if th

Internal architecture of microprocessor, Internal Architecture of Microproc...

Internal Architecture of Microprocessor : The architecture of 8086 provides a number of improvements over 8085 architecture. It supports a, a set of 16-bit registers ,16-bit AL

Comparison between 8086 and 8088, Comparison between 8086 and 8088 All ...

Comparison between 8086 and 8088 All the changes in 8088 above 8086 are indirectly or directly related to the 8-bit, 8085 compatible data and control bus interface. 1) The p

Assignment, 1. Write an assembly program that adds the elements in the odd ...

1. Write an assembly program that adds the elements in the odd indices of the following array. Use LOOP. What is the final value in the register? array1 DWORD 10, 20, 30, 40, 50, 6

Progframmw, write a programme the addition two 3*3 matrix and stored in fro...

write a programme the addition two 3*3 matrix and stored in from list

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd