Transcript Ch. I-1
CS2422 Basic Concepts Department of Computer Science National Tsing Hua University Assembly Language for IntelBased Computers, 5th Edition Kip Irvine Chapter 1: Basic Concepts Slides prepared by the author, with modification Revision date: June 3, 2006 (c) Pearson Education, 2006-2007. All rights reserved. You may modify and copy this slide show for your personal use, or for use in the classroom, as long as this copyright statement, the author's name, and the title are not changed. 當你在寫C程式時， 心中想像的電腦是長什麼樣子？ CPU, memory, I/O, Operations on variables, … A Model of Computer for C Program Counter i = i + j; xfloat = 1.0; if (A[0]==0) … Memory i, j, k; && CPU if + for xfloat, yfloat; A[0], A[1], … Typed storage 3 這個model和你實際跑程式的 電腦有很大的差別！ 為什麼你用這個C的model 所寫的程式能在你的電腦上執行？ 顯然有人為你做了轉換！ 我們已經知道... C program x = (a+b) * b Assembly program C compiler MOV ADD MUL MOV AX, a AX, b c x, AX 5 可見得組合語言程式更靠近 實際的電腦 組合語言程看到的電腦 是長什麼樣子? A Model of Computer for ASM MOV AX, a ADD AX, b MOV x, AX … CPU PC AX BX ALU Register Take the data stored in memory address a, and move it to register AX Take the data stored in memory address b, and add it to register AX Take the dataMemory stored in register AX, and move it to memory address x a 010100110010101 b 110010110001010 x 000000000010010 ... + - 7 組合語言程式/機器碼 看到的電腦， 離實際電腦還是有一段差距 e.g. 多核心、hyperthreading 當然仍需要再一或多層的轉換！ Computer Model of a Lower Layer PCSrc ID/EX 0 M u x 1 WB Control IF/ID EX/MEM M WB EX M MEM/WB WB Add Add Add result Instruction memory ALUSrc Read register 1 Read data 1 Read register 2 Registers Read Write data 2 register Write data MemtoReg Address Branch Shift left 2 MemWrite PC Instruction RegWrite 4 Zero ALU 0 M u x 1 ALU result Address Data memory Read data Write data Instruction 16 [15– 0] Instruction [20– 16] Instruction [15– 11] Sign extend 32 6 ALU control 0 M u x 1 MemRead ALUOp RegDst (from Computer Architecture textbook) 1 M u x 0 A Layered View of Computer i = i + j; if (A[0]==0) … Program Counter && CPU if i = i + j; xfloat = 1.0; if (A[0]==0) … i, j, k; xfloat, yfloat; + A[0], A[1], … for Memory ADD AX, b MOV x, AX … MOV AX, a ADD AX, b MOV x, AX … a 010100110010101 b 110010110001010 AX PC BX ... JX CPU Memory x 000000000010010 + PCSrc ID/EX 0 M u x 1 WB Control IF/ID EX/MEM M WB EX M MEM/WB WB Add Add Add result Instruction memory Branch Shift left 2 ALUSrc Read register 1 Read data 1 Read register 2 Registers Read Write data 2 register Write data MemtoReg Address MemWrite PC Instruction RegWrite 4 Zero ALU 0 M u x 1 ALU result Address Data memory Read data Write data Instruction [15– 0] Instruction [20– 16] Instruction [15– 11] 16 Sign extend 32 6 ALU control 0 M u x 1 ALUOp RegDst MemRead 1 M u x 0 xxxxxx xxxxx 10 What’s Next? Virtual machine concept (Sec. 1-2) Data representation (Sec. 1-3) 11 Virtual Machine Concept Purpose of this section: Understand the role of assembly language in a computer system Side product: The principle of layered abstraction for combating complexities, e.g. OSI 7-layer protocol 12 Virtual Machine Concept A layered abstraction of computers proposed by A. Tanenbaum Each layer provides an abstract computer, or virtual machine, to its upper layer Virtual machine: A hypothetical computer that can be constructed of either HW or SW What is a computer? High-Level Language Level 5 Assembly Language Level 4 Operating System Level 3 Instruction Set Architecture Level 2 Microarchitecture Level 1 Digital Logic Level 0 13 Simplest Model of Computers Instructions Program Input data Compute engine Memory Output data c.f., y = f(x) Layered abstraction: A computer consists of layers of such virtual machine abstractions 14 Why Layered Abstraction? Big idea: layered abstraction to combat complexities A strategy of divide-and-conquer Decompose a complex system into layers with well-defined interfaces Each layer is easier to manage and handle Only need to focus on a particular layer, e.g. to make it the best Also, it makes interaction clear Particularly if one layer is realized in hardware and the other in software 15 Layered Abstraction of Computer Each layer as a hypothetical computer, or virtual machine, that runs a programming language Can be programmed with the programming language to process inputs and outputs Instructions Compute engine Program written in Li can be mapped to that Li-1 by: Program Input data Memory Output data Interpretation: Li-1 program interprets and executes Li instructions one by one Translation: Li program is completely translated into Li-1 program, and runs on Li-1 machine 16 Layered Abstraction of Computer Program Counter && CPU if Memory i = i + j; xfloat = 1.0; if (A[0]==0) … i, j, k; xfloat, yfloat; + A[0], A[1], … for Memory ADD AX, b MOV x, AX … b 110010110001010 x 000000000010010 + - Li MOV AX, a ADD AX, b MOV x, AX … a 010100110010101 AX PC BX ... JX CPU Virtual Machine i = i + j; if (A[0]==0) … Li-1 PCSrc ID/EX 0 M u x 1 WB Control IF/ID EX/MEM M WB EX M MEM/WB WB Add Add Add result Instruction memory Branch Shift left 2 ALUSrc Read register 1 Read data 1 Read register 2 Registers Read Write data 2 register Write data MemtoReg Address MemWrite PC Instruction RegWrite 4 Zero ALU 0 M u x 1 ALU result Address Data memory Read data Write data Instruction [15– 0] Instruction [20– 16] Instruction [15– 11] 16 Sign extend 32 6 ALU control 0 M u x 1 ALUOp RegDst MemRead 1 M u x 0 xxxxxx xxxxx 17 Languages of Different Layers English: Display the sum of A times B plus C. C++: cout << (A * B + C); Assembly Language: mov eax,A mul B add eax,C call WriteInt Intel Machine Language: A1 00000000 F7 25 00000004 03 05 00000008 E8 00500000 18 High-Level Language Level 5 Application-oriented languages, e.g., C, C++, Java, Perl Written with certain programming model in mind Variables in storage Operators for operations Programs compiled into assembly language (Level 4) or interpreted by interpreters What kind of computer does C see? 19 Assembly Language Level 4 Instruction mnemonics that have a one-to-one correspondence to machine language Based on a view of machine: register organization, addressing, operand types and locations, functional units, … High-Level Language Calls functions written at the Assembly Language OS level (Level 3) Programs are translated into Operating System machine language (Level 2) Instruction Set What kind of computer does it see? Level 5 Level 4 Level 3 Architecture Level 2 Microarchitecture Level 1 Digital Logic Level 0 20 Operating System Level 3 Provides services to Level 4 programs as if it were a computer How about VMware? How about JVM? 21 Instruction Set Architecture Level 2 Known as conventional machine language Attributes of a computer as seen by assembly programmer, i.e. conceptual structure and functional behavior Organization of programmable storage Data types and data structures Instruction set and formats Addressing modes and data accessing Executed by Level 1 program (microarchitecture) 22 Microarchitecture Level 1 Can be described by register transfer language (RTL) Interprets conventional machine instructions (Level 2) Executed by digital hardware (Level 0) Register Control Signals Controller clock Memory N Z IR ALU PC 23 Digital Logic Level 0 CPU, constructed from digital logic gates System bus Memory 24 What’s Next? Virtual machine concept (Sec. 1-2) Data representation (Sec. 1-3) 25 Data Representation Purpose of this section Assembly program often needs to process data, and manage data storage and memory locations need to know data representation and storage Binary numbers: translating between binary and decimal Binary addition Integer storage sizes Hexadecimal integers: translating between decimal and hex.; hex. subtraction Signed integers: binary subtraction Character storage 26 Integer Storage Sizes Standard sizes: byte word doubleword 8 16 32 quadword 64 Why unsigned numbers? What is the largest unsigned integer that may be stored in 20 bits? 27 Signed Integers The highest bit indicates the sign 1 = negative, 0 = positive sign bit 1 1 1 1 0 1 1 0 0 0 0 0 1 0 1 0 Negative Positive If the highest digit of a hexadecimal integer is > 7, the value is negative. Examples: 8A, C5, A2, 9D 28 Forming Two's Complement Negative numbers are stored in two's complement notation Complement (reverse) each bit Add 1 Why? Note that 00000001 + 11111111 = 00000000 29 Binary Subtraction When subtracting A – B, convert B to its two's complement Add A to (–B) 1100 – 0011 1100 + 1101 1001 Practice: Subtract 0101 from 1001 30 Ranges of Signed Integers The highest bit is reserved for the sign This limits the range: What is the largest positive value that may be stored in 20 bits? 31 Character Storage Character sets Standard ASCII (0 – 127) ‒ American Standard Code for Information Interchange Extended ASCII (0 – 255) ANSI (0 – 255) ‒ American National Standard Institute Null-terminated string Unicode (0 – 65,535): UTF-8. UTF-16, UTF-32 Array of characters followed by a null byte Using the ASCII table back inside cover of book 32 Chapter Summary Assembly language helps you learn how software is constructed at the lowest levels Assembly language has a one-to-one relationship with machine language Each layer in a computer's architecture is an abstraction of a machine layers can be hardware or software Assembly programs and assemblers often need to manage data storage in memory, and it is important to understand data representation and size of different data 33