kaspersky internet security 2018 system requirements

Want to know kaspersky internet security 2018 system requirements? we have a huge selection of kaspersky internet security 2018 system requirements information on alibabacloud.com

2018-2019-1 20165322 "The foundation of Information security system Design" buffer Overflow vulnerability experiment

enable the compiled executable document to be debugged with GDB New exploit.c, code below, \x?? \x?? \x?? \x?? Need to add shellcode to the address stored in memory because the location can overwrite the return address just after an overflow occurs. We want to get shellcode in-memory address, enter commands gdb stack anddisass main According to strcpy(buffer + 100,shellcode) the statement, we calculate shellcode the address as0xffffd350(十六进制) + 0x64(100的十六进制) = 0xffffd3b4(十六进制) Mo

2018-2019-1 20165221 summary of the fifth week of Information Security System Design Basics

lower layer. Due to locality, writing back can significantly reduce bus traffic.The disadvantage is increased complexity.Another problem is how to deal with write Miss.Write allocation: load the corresponding lower-level block to the cache, and then update the cache block. The disadvantage is that each Miss will cause a block to be transferred from the lower layer to the cache.Non-write allocation: Avoid high-speed caching and write the word directly to the lower layer.Performance impact of hig

2018-2019-1 20165320 "The foundation of Information security system Design" buffer Overflow vulnerability experiment

exploit.c file to compile.6. Running the programAfter the address protection mechanism has been modified, the run is unsuccessful and the segment error is displayed. Experimental HarvestFeel according to the steps of the experiment step by step, really can get the corresponding results, but the whole experiment is a walk a process, or do not understand the relevant memory overflow of the specific process, as well as the relevant attack ideas, their level or not standards, there is a long way

2018-2019-1 20165214 "Information Security system Design Fundamentals" Third Week study summary

20165214 2018-2017-1 "Information Security system Design Fundamentals" The third week study summary of learning Contents1. In fact, the GCC command invokes a series of programs that convert the source code into executable code.2, the actual realization of the memory system is to combine multiple hardware memory and ope

2018-2019 20165203 summary of the fifth week of Information Security System Design Basics

2018-2019 20165203 Information Security System Design Basics Chapter 5 study summary teaching material content summary important concepts Storage System: a hierarchy of storage devices with different capacity costs and access time. Cache memory ------ buffer area for data and instructions in primary storage ------ d

2018-2019-1 20165302 "Information Security system Design Fundamentals" Third Week study summary

1. Machine-Level Code(1) Two kinds of abstract Defines the format and behavior of machine-level programs by ISA The memory address used by the machine-level program is the virtual address 2. Data format3. Operand designator4. Press in and eject stack data Follow the principle of first in and out Push Press in, pop delete Pushq press four words into the stack popq four words pop-up stack 5. Arithmetic and logical operations LEAQ Load Valid address INC plus a D

2018-2019-1 20165330 "Information Security system Design Fundamentals" Fourth Week study summary

non-gate or non-gate HCL integer Expression Case Expression Format: [ select 1: expr 1 select 2: expr 2 . select k: expr k ] Set Relationship:iexp in{ iexp1,iexp2,...iexpk } Arithmetic/logic unit (ALU) Sequential implementation of Y86-64 Organize the processing into stages Value fetch--> decoding decode--> performing execute--> memory--> writeback write back write back--> update PC update SEQ

2018-2019-1 20165304 "Information Security system Design Fundamentals" Fourth Week study summary

week's exam error summary 1.The following jump commands are related to ZF ()A. jmpB. JeC. jsD. JaE. JBF. JbeAnalytical:2.Assuming that the function of the C-expression T=a+b is completed with the add instruction, the correct statement about the condition Code Register is ()A. If t==0, then zf=1B. If tC. If tD. if (aE. if (aF. LEAQ directive does not affect the condition code registerG. CMP directives do not affect the condition code registerAnalysis: Textbook p135ZF: 0 logo. The result of the r

2018-2019-1 20165234 "Information Security system Design Fundamentals" Fourth Week study summary

I. Learning Objectives Understanding the role of ISA abstraction Master Isa, and be able to learn other architecture extrapolate Understanding the pipeline and how it is implemented Second, the Learning content y86-64 directive MOVQ directive IRMOVQ rrmovq mrmovq RMMOVQ Four integer manipulation instructions Addq,subq,andq,xorq only the Register data 7 Jump Instructions Cmovle cmovl cmove cmovne cmovge CMOVG The call command returns the address to the stack, and then j

2018-2019-1 20165302 "Information Security system Design Fundamentals" Fourth Week study summary

1.Y86-64 Instruction Set architecture①Y86-64 directive MOVQ directive IRMOVQ rrmovq mrmovq RMMOVQ Four integer manipulation instructions Addq,subq,andq,xorq only the Register data 7 Jump Instructions Cmovle cmovl cmove cmovne cmovge CMOVG The call command returns the address to the stack, and then jumps to the destination address, and the RET instruction returns from such calls Pushq and POPQ instructions are implemented into the stack and out of the stack Execution of Halt

2018-2019-1 20165228 "Information Security system Design Fundamentals" Third Week study summary

2018-2019-1 20165228 "Fundamentals of Information Security system design" the third week of learning summary of the Learning Content Summary program machine-level representation: Two important abstractions of computer systems ISA (Instruction set architecture): Instruction set architecture, machine-level program format and behavior. Defines the format of

2018-2019-1 20165336 "Information Security system Design Fundamentals" Fourth Week study summary

2018-2019-1 20165336 "Information Security system Design Fundamentals" Fourth week study summary 1. The knowledge points learned in the textbook The states that are visible to programmers (assembler programmers, compilers, and so on) in y86-64 include program registers, condition codes, program states, program counters (PCS), memory Y86-64 15 Program regist

2018-2019 20165219 summary of the fifth week of Information Security System Design Basics

2018-2019 20165219 Summary of storage technology in week 5 of Information Security System Design Basics Random Access Memory: Comparison of SRAM and DRAM, traditional DRAM and enhanced dram Enhanced DRAM: Fast page mode DRAM: allows continuous access to the same row to get services directly from the row buffer. Extended Data Output DRAM: Allows CAS signals to

2018-2019-1 20165228 "Information Security system Design Fundamentals" Second week study summary

2018-2019-1 20165228 "The foundation of Information security system design" The second week study summary textbook learning content Summary information = bit + context Unsigned encoding: Represents a number greater than or equal to zero based on the traditional binary notation Complement coding: represents the most common way to sign a certificate, a number

2018-2019-1 20165329 "Information Security system Design Fundamentals" 4th Week Study Summary

2018-2019-1 20165329 "Information Security system Design Fundamentals" 4th Week Study SummarySummary of learning contents of textbook Y86-64 directive: The y86-64 instruction is a subset of the x86-84 instruction set. It includes only 8-byte integer operations. There are 4 integer operations directives: ADDQ, SUBQ, ANDQ, and Xorq. There are 7 jump commands: j

2018-2019 20165227 "Fundamentals of Information Security system Design" Third week study summary

2018-2019 20165227 "Information Security system Design Fundamentals" The third week to learn to summarize learning objectives Understanding the concept of reverse Master X86 compilation base, able to read (reverse) Assembly code Understanding ISA (Instruction set architecture) Understand the concept of function call stack frames and can debug

2018-2019-1 20165206 "Foundation of Information Security System Design" 2nd weekly study summary

-2018-2019-1 20165206 "basis of information security system design" weekly learning Summary-textbook learning content Summary-hexadecimal conversion and bitwise calculation: Converts a hexadecimal parameter to a binary representation and performs binary operations. And, or, not, or, in bitwise operations, corresponds to , |, ~ in C ,~ ^-Logical operation: Only 0

2018-2019-1 20165314 summary of the fifth week of Information Security System Design Basics

2018-2019-1 20165314 summary of the fifth week of Information Security System Design Basics SRAM and DRAM are both easy to lose, but they have different structures, different speeds, and faster access speed. The prom can only be programmed once. Programs stored in ROM devices are usually called firmware. The EEPROM can erase the information on the chip with

2018-2019-1 20165333 "Information Security system Design Fundamentals" Third Week study summary

2018-2019-1 20165333 "Information Security system Design Fundamentals" Third Week study summarySummary of learning contents of textbookThe machine-level representation of the program:Two important abstractions of computer systemsISA (Instruction set architecture): Instruction set architecture, machine-level program format and behavior. Defines the format of the p

2018-2019-1 20165323 "Foundation of Information Security System Design" 5th weekly study summary

selection: determine where the desired word starts in the block. Associated High-speed cache:1. group selection in group-connected high-speed cache: Same as group selection in direct ing high-speed cache, group index bit Identification Group.2. Row matching and word selection in the group-connected high-speed cache: each group is considered as a small memory associated with a (Key, value) pair array,3. Input key to return the value in the corresponding array. The cache must search for each row

Total Pages: 3 1 2 3 Go to: Go

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.