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1、计算机导论考前辅导助教刘阳学习方法应试方法 认真对待作业题，作业题就是考试题 老师上课讲的内容就是考试内容（不必死扣书本） 考试不交白卷，越不会越多写（相关公式、概念、自己的想法） 大学考试考的是记忆力，请珍惜自己的大脑 神通广大的打印店（复习资料、历年试卷）Chapter 1 IntroductionComputer Is a programmable data processor that accepts input data and programs and outputs data.ComputerSystem Solve problems and interact with thei
2、r environment. They consist of Devices, programs, and data.Program Is a set of instructions executed sequentially that tells the computer what to do with data.Chapter 1 IntroductionVonNeumannModellFourSubsystemsMemory（内存） Arithmetic Logic Unit（ALU ，算术逻辑单元） Control Unit（控制器） Input/Output（输入/输出设备）lSto
3、redProgramConceptlSequentialExecutionOfInstructionsChapter 2 Data RepresentationDataTypeslText BytelNumber ASCIIlImage Bitmap，Vector，Pixel，ResolutionlAudio Digital，AnaloglVideo FrameChapter 2 Data RepresentationTextlBit（位） A bit (binary digit) is the smallest unit of data that can be stored in a com
4、puter，it is either 0 or 1. lBitPattern（位组合格式） A bit pattern is a sequence, or as it is sometimes called, a string of bits that can represent a symbol. lByte（字节） A bit pattern of length 8 is called a byte. Chapter 2 Data RepresentationNumberlASCII American Standard Code for Information Interchange (A
5、SCII)（美国信息交换标准代码）. This code uses 7 bits for each symbol. This means 128 different symbols can be defined by this code.Chapter 2 Data RepresentationImagelBitmap Graphic Pixel: Picture elementsResolution:The size of the pixelChapter 2 Data RepresentationImagelVector Graphic An image is decomposed int
6、o a combination of curves and lines. Each curve or line is represented by a mathematical formula.Chapter 3 Number RepresentationConceptsDecimalsystem(十十进制制) based on 10，0-9；Binarysystem(二二进制制) based on 2，0-1；Octalnotation(八八进制制) based on 8，0-7；Hexadecimalnotation(十六十六进制制) based on 16 ，0-9，A-F。Chapte
7、r 3 Number RepresentationConversionDecimalBinaryHexadecimalnotationOctalnotationChapter 3 Number RepresentationIntegerRepresentationUnsignedInteger（无符号整数）（无符号整数） Overflow（溢出）（溢出）Sign-and-MagnitudeFormat（原码）（原码） +0 00000000 -0 10000000Range:0.(2N-1)Range:-(2N-1-1)+(2N-1-1)Chapter 3 Number Representat
8、ionIntegerRepresentationOnesComplementFormat（反码（反码） +0 00000000 -0 11111111TwosComplementFormat（补码）（补码）Range:-2N-1+(2N-1-1)Range:-(2N-1-1)+(2N-1-1)Chapter 3 Number RepresentationExcessSystemMagicNumberIs normally (2N-1) or (2N-1-1), where N is the bit allocation.Example:Represent 25 in Excess_127 us
9、ing an 8-bit allocation. -25 + 127 102 1100110 01100110Chapter 3 Number RepresentationFloating-PointRepresentation 1. Convert the integerpartto binary. 2. Convert the fraction to binary. 3. Put a decimal point between the two parts.Example: Transform the fraction 0.875 to binary0.875 0.111Chapter 3
10、Number RepresentationNormalization Example: Show the representation of the normalized number + 26 x 6 + 127 133 10000101010000101010000101Signexponent mantissa Original NumberOriginal Number-+1010001.1101-111.000011+0.00000111001-001110011MoveMove- 6 26 3 Normalized - +26 x 1.01000111001 -22 x 1.110
11、00011 +2-6 x 1.11001 -2-3 x 1.110011Chapter 4 Operations on BitsArithmeticOperationsExample1:Solution: Add two numbers in twos complement representation: (+24) + (-17) (+7)Carry 1 1 1 1 10 0 0 1 1 0 0 0 + +1 1 1 0 1 1 1 1-Result 0 0 0 0 0 1 1 1 +7Chapter 4 Operations on BitsArithmeticOperationsExamp
12、le2:Solution:Add two numbers in twos complement representation: (+127) + (+3) (+130)Carry 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 + + 0 0 0 0 0 0 1 1 -Result 1 0 0 0 0 0 1 0 -126(Error) Anoverflowhasoccurred.Chapter 4 Operations on BitsArithmeticOperationsExample3:Solution: Subtract 62 from 101 in twos comple
13、ment: (+101) - (+62) (+101) + (-62)Carry 1 10 1 1 0 0 1 0 1 + +1 1 0 0 0 0 1 0-Result 0 0 1 0 0 1 1 1 39The leftmost carry is discarded.Chapter 4 Operations on BitsLogicalOperationsChapter 4 Operations on BitsLogicalOperationsExample1:Solution:Use the NOT operator on the bit pattern 10011000.Target
14、Target 1001100010011000 NOTNOT - -Result Result 0110011101100111Chapter 4 Operations on BitsLogicalOperationsExample2:Solution:Use the AND operator on bit patterns 10011000 and 00110101.Target Target 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 0 ANDAND 0 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 - -Result 0 0 0 1 0 0 0 0Resu
15、lt 0 0 0 1 0 0 0 0Chapter 4 Operations on BitsLogicalOperationsExample3:Solution:Use the OR operator on bit patterns 10011000 and 00110101.Target Target 1001100010011000 OROR 0011010100110101 -Result Result 1011110110111101Chapter 4 Operations on BitsLogicalOperationsExample4:Solution:Use the XOR op
16、erator on bit patterns 10011000 and 00110101.Target Target 1001100010011000 XORXOR 0011010100110101 -Result Result 1010110110101101Chapter 4 Operations on BitsLogicalOperationsExample5:Solution:Use a mask to unset (clear) the 5 leftmost bits of a pattern. Test the mask with the pattern 10100110.The
17、mask is The mask is 0000011100000111. .Target Target 1010011010100110 ANDMask Mask 0000011100000111 -Result Result 0000011000000110Chapter 4 Operations on BitsShiftOperations Shift operations can only be used when a pattern represents an unsigned number.左移：左移：*2右移：右移：/2Chapter 5 Computer Organizatio
18、nMainMemoryUnitUnit-Kilobyte(K)Megabyte(M)Gigabyte(G)Terabyte(T)Petabyte(P)Exabyte(E)Exact Number of bytesExact Number of bytes-210 bytes220 bytes230 bytes240 bytes250 bytes260 bytesApproximationApproximation-103 bytes106 bytes109 bytes1012 bytes1015 bytes1018 bytesKB Kilobyte：千字节，1024字节MB Megabyte：
19、兆字节GB Gigabyte：十亿字节TB Terabyte：1000吉千兆字节GB；兆兆位Chapter 5 Computer OrganizationMainMemoryExample:Solution:A computer has 32 MB (megabytes) of memory. How many bits are needed to address any single byte in memory?The memory address space is 32 MB, or 2The memory address space is 32 MB, or 22525 (2 (25
20、5 x 2 x 22020). ). This means you needThis means you needloglog2 2 2 22525 or 25 bits, to address each byte. or 25 bits, to address each byte.Sign-and-magnitude, ones complement and twos complement 32-bit IEEE format 可能的未来学神个数：1 2 个成绩：1 120名之间特点： 专业技能极强，精通各种编程语言 参加过ACM竞赛、机器人大赛、数学建模大赛等取得好成绩毕业去向： 知名IT企业（谷歌中国、阿里巴巴等）年薪 30w 50w 保送国内名校可能的未来学霸个数：20个左右成绩：1 30名之间特点： 学习成绩优异 外语水平好（托福、GRE）毕业去向： 保送国内名校 出国留学（世界名校） 个别选择直接就业可能的未来学子个数：80个左右成绩：20 100名之间特点： 大学过的很Happy 毕业去向： 考研 就业 8w 15w 出国（自主） 可能的未来学渣个数：5个左右成绩：80 120名之间特点： 沉迷游戏，挂科严重毕业去向： 大五 大六 就业（没有学位证，只有中小私企会要） 青年湖底可能的未来 ？结束语 谢谢！
