STM32 reverse entry

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stm32 Reverse entry _ The blog of Hetian Wangan Laboratory -CSDN Blog

Pay attention to our

One 、 Preface
Today's learning record

Two 、 Reappear

1、SCTF 2020 Password Lock

Reference link ：https://xuanxuanblingbling.github.io/iot/2020/07/08/stm32/

Title Description ：

This is a STM32F103C8T6 MCU Password lock It has 4 A button , Respectively 1, 2, 3, 4. They correspond to each other GPIO_PA1, GPIO_PA2, GPIO_PA3, GPIO_PA4 flag1 The format is SCTF{ Correct key password } Enter the correct password , It will pass the serial port (PA9–TX) send out flag2

Their thinking ：

The attachment of the title gives a Intel hex file , And given the chip information, we can determine the memory layout of the program and the correspondence between peripheral registers and memory . The key to reverse is to understand the meaning of program code , Next, we will analyze this step by step hex file .

1. hex File structure

Intel hex The file format consists of plain text , It contains information such as the loading address and entry address of the program , Reading this information can help us quickly locate the starting entry of the program instead of ida To configure .

• Read Intel hex file

We can use a text editor to open the title attachment , The key information is as follows ：

:020000040800F2
...
...
:04000005080000ED02
:00000001FF

• The program loading address is 0x08000000

• The program entry address is 0x080000ED

• Procedure to :00000001FF ending

• The rest is all file data

2、 Memory layout

Find the website of the chip manual ：https://www.alldatasheet.com/ Inside we can find STM32F103C8T6 Manuals , The first page finds some information we need

• Flash memory：32-to-128 Kbytes
• SRAM：6-to-20 Kbytes

31 page Memory Map It can let us more intuitively understand the detailed layout of memory

To sum up, we got the complete memory layout information of the program ：

• Flash Memory: 0x8000000 ~ 0x801FFFF (128K)
• SRAM: 0x20000000 ~ 0x20004FFF (20K)
• Peripherals: 0x40000000 ~ 0x40023400

3、IDA analysis

After the analysis just now, we know the memory layout of the program , among Flash In addition to containing code , And interrupt vector table .Periphers The register in the segment is that we need to have a general understanding of alignment in the reverse process . And for hex Through the analysis of the file, we learned that in addition to the loading address and entry address , Everything else is gone hex In file , So we need to manually configure these memory layout information to tell IDA How to identify . open ida Tools , According to the manual just now, we can find that the chip is arm32 Armv7-M framework , Select the configuration as shown in the following figure, and then click ok

You can see that some functions can be recognized , among start The address of the function is analyzed with us hex The program entry address found in the file structure is the same .

If hex The entry address information is not given in the file. We can also find RESET Interrupt handler function to determine program entry function . among RESET The address of the interrupt function can be in STM32 Chinese Reference Manual V10.pdf Find relevant information in

Reference resources STM32 The location of the interrupt vector table 、 Redirect We can see in the interrupt vector table RESET The address of 0x8000004 The address is fixed , What is changeable is the loading address of the program . We jump to 0x8000004 On this address , Press keyboard D The key divides the above data into 4 Byte form found reset The address for 0x8000101

Jump to RESET Interrupt handling function , There are two jumps . Jump to for the first time nullsub_1 Up and put the address of the next instruction into LR register ,nullsub_1 The function is to jump back LR Address in register , So the first jump is meaningless . The second jump is ours start Address , So you can use this method to locate the entry address of the program . 

You can find this program by following the entry address all the time main Where is the function , But after entering, you can find a large red mark on the left , Observing these red areas is actually IDA Unrecognized address , That is, the memory segments we need to add before analyzing the memory layout .

We are IDA New China Segment, As shown in the figure below ： 

At this time, as long as we click again F5 You can turn these red signs into memory for normal recognition

4、 Fix interrupt vector table

Use IDApython Recover the information before the program entry address

for i in range(0x8000000,0x80000eb,1): 
 del_items(i)
for i in range(0x8000000,0x80000eb,4): 
 create_dword(i)

After the repair, we observe the address inside , You will find many duplicate addresses 0x800016D, You will find that there is no function defined in these addresses . Continue to check the interrupt vector table and you will find the following different memory addresses

Reference resources STM32 Chinese Reference Manual V10.pdf We can find these in EXTI Address of the interrupt handling function

• stm32-EXTI

Follow up these function addresses and you will find IDA It is not recognized as a function , So let's first press... At the starting address of the function P key , Then disassemble to see these interrupt handling functions , Here I use EXTI_4 Take the interrupt handling function of as an example to briefly introduce the functions of these interrupt handling functions .

int EXTI_4()
{
  int result; // r0
 
  EXTI_LINE = 16;
  switch ( sum )
  {
    case 1:
      unk_20000006 = 116;
      return sum++ + 1;
    case 2:
      unk_20000010 = 95;
      return sum++ + 1;
    case 4:
      unk_2000000E = unk_20000001;
      return sum++ + 1;
    default:
      result = 0;
      sum = 0;
      break;
  }
  return result;
}

The program starts with an interrupt / Event line ,EXTI_4 The interrupt / The event line is 0x10, Then use a piece of memory ( It's written here sum) As the storage location of the accumulated number . We can see when sum The value of 1、2、4 when sum It's worth it +1, If not, it will start again . So we can judge 1、2、4 Namely EXTI_4 The number of digits appearing in the password , Similarly, the other three buttons are the same , Through these sequences, we can get the final flag by flag{1442413}, And we can find in main In the function, the program first simulated the input of a password , By comparing the above value with EXTI_LINE We can also get flag value .

2、2021 HWS Enter the camp -STM32

After the introduction of the previous question, let's practice another question , open IDA Type selection segment arm32, Schema selection ARMv7-M framework .

Next, set the loading address and reading address of the program to 0x8000000, The loading address refers to IDA What is the analysis address loaded , and Input File It refers to where the firmware should be loaded , After setting, we click OK Complete set . 

After entering, you will find IDA No function was recognized , Don't worry, we can locate reset Interrupt processing to find main Position of function . take 0x8000004 The byte at the address becomes 4 byte , obtain reset Interrupt handler address 0x8000101. 

 

You can find that the end of the address is an odd digit , stay arm This represents thumb Pattern . We can do it in 0x8000101 Address press C Key to generate code  

Follow the program all the time and we will find main Function location sub_80003C0, And found the function that needs to be reversed sub_8000314

_BYTE *sub_8000314()
{
  _BYTE *v0; // r4
  char *v1; // r5
  int v2; // r6
  char v3; // t1
 
  v0 = (_BYTE *)sub_80003F0(48);
  v1 = &byte_8000344;
  v2 = 0;
  while ( v2++ != 0 )
  {
    v3 = *v1++;
    *v0++ = (v3 ^ 0x1E) + 3;
    sub_8000124(v1);
  }
  return v0;
}

  Write a problem-solving script to get flag value

li = [0x7D, 0x77, 0x40, 0x7A, 0x66, 0x30, 0x2A, 0x2F, 0x28, 0x40, 0x7E, 0x30, 0x33, 0x34, 0x2C, 0x2E, 0x2B, 0x28, 0x34, 0x30, 0x30, 0x7C, 0x41, 0x34, 0x28, 0x33, 0x7E, 0x30, 0x34, 0x33, 0x33, 0x30, 0x7E, 0x2F, 0x31, 0x2A, 0x41, 0x7F, 0x2F, 0x28, 0x2E, 0x64]
print(''.join(chr((i ^ 0x1E) + 3) for i in li))

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