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

Code for reading flow & generating serial output, Assembly Code for Reading...

Assembly Code for Reading Flow & Generating Serial Output The timer is timer 1 is set for the baud rate 9600, as the crystal used is of 11.0592 Hz.  Then the timer 1 is starte

Program to accept the input from user, Write an assembly language program t...

Write an assembly language program that will: accept keyboard input of a positive integer value N; compute the sum S= 1+ 2 + 3 + ... + N; print (output) the computed su

General bus operation-microprocessor, General Bus Operation The 8086 ha...

General Bus Operation The 8086 has a joined data and address bus commonly referred to as a time multiplexed address and data bus. The major reason behind  multiplexing address

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

Hold response sequence-microprocesssor, Hold Response Sequence The HOLD...

Hold Response Sequence The HOLD pin is examined at leading edge of each clock pulse. If it is received active line by the processor before T4 of the earlier cycle/during the T1

#title, how i can write a program to divide 2 numbers

how i can write a program to divide 2 numbers

Read architecture:look through-microprocessor, Read Architecture: Look Thro...

Read Architecture: Look Through Main memory that located is conflicting the system interface. The least concerning feature of this cache unit is that it remain between the proc

Internal hardware-interrupts-microprocessor, Internal Hardware-Interrupts ...

Internal Hardware-Interrupts Internal hardware-interrupts are the outcome of sure situations that occur during the execution of a program, for example. Divide by 0. The interru

Eax and ax register, MyLocation SDWORD 14 TheTest        SDWORD 8     mov ...

MyLocation SDWORD 14 TheTest        SDWORD 8     mov    eax,MyLocation     mov    ebx,TheTest     neg     eax,ebx     sub     eax,ebx Show exactly what lives in eax after executi

Help with homework , I need help with assembly programing

I need help with assembly programing

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