User Account

All Pages

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

Sequence of interrupts, As an instance of the normal priority mode, imagine...

As an instance of the normal priority mode, imagine that initially AEOI is equal to 0 and all the ISR and IMR bits are clear. Also consider that, as shown in given figure, requests

Introduction to evaluation, This unit introduces the topic of evaluating in...

This unit introduces the topic of evaluating interactive products. It is a short unit as evaluation is discussed in more detail in Block 4. Its brevity should give you additional t

Addressing mode of 8086-microprocessor, Addressing mode of 8086 : Addre...

Addressing mode of 8086 : Addressing mode specify a way of locating operands or data. Depending on the data types used the memory  addressing  modes and in the instruction  ,

Memory interface-microprocessor, Memory Interface              ...

Memory Interface                                                                  Figure: Memory Modulation design The memory of a computer contain of number of memo

Dma-how dma works-microprocessor, DMA DMA stands for Direct Memory ...

DMA DMA stands for Direct Memory Access It is uses same Address/Data lines on ISA bus It controls the ISA bus instead of the processor ("bus master") Floppy

Write program that will generate array of ten random number, 1. Write a pro...

1. Write a program that will generate an array of ten random 32-bit integers, and that will  display on the monitor the numbers followed by either the words " has the fourth bit se

Matlab?., Hello, I just want to know how much would it cost for you to deve...

Hello, I just want to know how much would it cost for you to develop , debug and test a program in matlab to solve a system of equations with gauss elimination with partial pivotin

Adc-arithmetic instruction-microprocessor, ADC: Add with Carry:- This instr...

ADC: Add with Carry:- This instruction performs the similar operation a like ADD instruction, but adds the carry flag bit (which might be set as a result of the previous calculatio

Intel''s 8237 dma controller-microprocessor, Intel's 8237 DMA controller : ...

Intel's 8237 DMA controller : 1) The 8237 contain 4 independent I/O channels 2) It contains 27 registers, 7 of which are system-wide registers and 5 for each channel. 3)

Login system, a pseudo-code to add username and password combination up to ...

a pseudo-code to add username and password combination up to a limit of 10

Write Your Message!

Captcha
Order Assignment

Featured Services

Order Now

Popular Subjects

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!

Submit Assignment

Academics

Major Subjects

Majors

Get In Touch

whatsapp: +91-977-207-8620

Phone: +91-977-207-8620

Email: [email protected]

TERMS & POLICIES

HELP & SUPPORT

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