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Computer Organization

This chapter provides an overview of memory location and addresses in computer organization. It covers topics such as memory organization, word length, byte addressability, word alignment, and accessing numbers, characters, and strings.

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Computer Organization

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  1. Chapter 2: Machine Instructions and Programs Computer Organization Department of CSE, SSE Mukka www.bookspar.com | Website for students | VTU NOTES

  2. Memory Location and Addresses • In memory of computer, there are • Number and character operands • Instructions • Memory consists of • Millions of storage cells, • Each cell can hold a bit ( 0 or 1 ) of information • So each bit can hold a very small amount of information • Memory is organized so that a group of n bits can be stored or retrieved in a single, basic operation • Each group of n bits is referred to as word • n is called as wordlength www.bookspar.com | Website for students | VTU NOTES

  3. n bits first word second word • • • i th word • • • last word Memory words. www.bookspar.com | Website for students | VTU NOTES

  4. Contd… • Range of word length of modern computers • 16 to 64 bits • Group of 8 bits referred to a byte • If the word length is 32 bits • A single word can store a 32-bit 2’s complement number • Or • Four ASCII characters • As shown in the figure www.bookspar.com | Website for students | VTU NOTES

  5. 32 bits • • • b b b b 31 30 1 0 for positive numbers Sign bit: b = 0 31 for negative numbers b = 1 31 (a) A signed integer 8 bits 8 bits 8 bits 8 bits ASCII ASCII ASCII ASCII character character character character (b) Four characters Examples of encoded information in a 32-bit word. www.bookspar.com | Website for students | VTU NOTES

  6. Addresses • Access to memory to store or retrieve a single item of info ( either a byte or a word ) requires distinct names or addresses for each item location • normally used addresses => 0 to 2k -1 as addresses of successive memory locations • The 2k addresses constitute address space of the computer • 24-bit address generates an address space of 224 locations = 16M locations • 32-bit ? www.bookspar.com | Website for students | VTU NOTES

  7. Byte Addressability • A byte is always 8-bit • Word length ranges from 16 to 64 bits. • Impractical to assign distinct addresses to individual bit locations • Most practical? • Byte-addressable memory • Byte locations have addresses 0,1,2,.. • If the word length of the machine is 32-bits, then successive words are located at addresses ? www.bookspar.com | Website for students | VTU NOTES

  8. Big-Endian and Little-Endian Assignments • 2 ways byte addresses can be assigned across words • Big-endian when • Lower byte addresses for more significant bytes (leftmost bytes ) of the word. • Little-endian when • Lower bytes addresses used for less significant bytes (right most bytes ) of the word. • In both cases 0,4,8 .. are taken as addresses of successive words in the memory www.bookspar.com | Website for students | VTU NOTES

  9. W ord address Byte address Byte address 0 0 1 2 3 0 3 2 1 0 4 4 5 6 7 4 7 6 5 4 • • • • • • k k k k k k k k k k 2 - 4 2 - 4 2 - 3 2 - 2 2 - 1 2 - 4 2 - 1 2 - 2 2 - 3 2 - 4 (a) Big-endian assignment (b) Little-endian assignment Byte and word addressing. www.bookspar.com | Website for students | VTU NOTES

  10. Word Alignment • Words are said to be aligned in memory if they begin at a byte address that is a multiple of the number of bytes in a word • Eg : if wordlength is 16bits, aligned words begin at byte addresses 0,2,4,…. • If word length is 32 bits, aligned words begin at 0,4,8 etc… • If the words don’t begin at byte address that is a multiple of no of bytes in the word, then words are said to have unaligned addresses www.bookspar.com | Website for students | VTU NOTES

  11. Accessing numbers, characters and strings • Number • By its word address as it usually occupies one word • Character • By byte address • Strings • They are of variable length • Beginning of the string by giving the beginning byte address which contains first character • Successive bytes contains successive characters • Termination? • Either by a special control character • Or a separate memory word location/ register containing a number indicating the string length www.bookspar.com | Website for students | VTU NOTES

  12. Problems • Given that a memory has 32-bit address & is byte-addressable, what is the size of the memory(in bytes)? • Given that a memory has 24-bit address & is word-addressable with a word length of 32 bits, what is the size of the memory(in bytes)? • Given that a memory has 16-bit address and is byte addressable. Word length is 32 bits. How many words can we store in such a memory? www.bookspar.com | Website for students | VTU NOTES

  13. Answers • 4GB • 64MB • 16K words www.bookspar.com | Website for students | VTU NOTES

  14. Memory Operations • Both program instructions and data operands are stored in memory • To execute an instruction • Processor control circuits must cause the word(s) containing the instruction to be transferred from the memory to the processor • Operands and results must also be moved between memory and the processor • Two basic operations involving the memory • Load ( or Read or Fetch ) • Store ( or Write ) www.bookspar.com | Website for students | VTU NOTES

  15. Memory Operations – LOAD operation • Load transfers a copy of the contents of a specific memory location to the processor • The memory contents remain unchanged • To start a load operation • Processes sends the address of the desired location to memory • Request that its contents be read • The memory reads the data and sends to the processor www.bookspar.com | Website for students | VTU NOTES

  16. Memory operations -Store operation • Store operation transfers an item of information from the processor to a specific memory location • Destroys the former contents of that memory location • Processor needs to send the address of the desired memory location and also the data to be written into that location www.bookspar.com | Website for students | VTU NOTES

  17. Instructions and Instruction Sequencing • Tasks that are carried out by a computer consists of a sequence of small steps • Eg., add two numbers, test for a particular condition, read a character from keyboard, display a character on screen • A computer must have instructions capable of performing four types of operations • Data transfers between the memory and the processor registers • Arithmetic and logic operations on data • Program sequencing and control • I/O transfers www.bookspar.com | Website for students | VTU NOTES

  18. Register transfer notation • Used to describe transfer of information from one location in the computer to another • Possible locations are memory locations, processor registers, and registers in the I/O subsystem • To identify a location • Symbolic name standing for its hardware binary address • Eg., name of memory locations – LOC,A,VAR2 etc • Name of registers – R0,R5 • Name of I/O registers – DATAIN • Contents of a location denoted by placing square brackets around the name • R1  [LOC] means contents of memory location LOC are transferred into processor register R1 • R3  [R1] + [R2] ? www.bookspar.com | Website for students | VTU NOTES

  19. Assembly language notation • Used to represent machine instructions and programs • Example • To perform the data transfer R1 [LOC] the statement is Move LOC,R1 • Old contents of register R1 are overwritten but contents of LOC unchanged • R3  [R1] + [R2] is denoted by statement ? www.bookspar.com | Website for students | VTU NOTES

  20. Basic Instruction Types • A high level language program command C = A + B adds the contents of variables A and B and stores the result in variable C • After compilation, the three variable A,B and C are assigned distinct locations in memory • The contents of these locations represent the value of the three variables • The action is C  [A] + [B] • Contents of A and B locations are fetched from memory and transferred into processor where the computation is performed • Result is then sent back to the location C www.bookspar.com | Website for students | VTU NOTES

  21. Accomplish C=A+B using Single Machine Instruction • The instruction contains memory addresses of 3 operands – A,B and C • Add A,B,C • A and B are source operands • C is destination operand • Add is the operation to be performed • Operation Source1,Source2, Destination • If k bits are needed to specify memory address of each operand • In addition to the bits needed to denote Add operation we need 3k bits more • For a modern processor with 32-bit address space,a 3-address instruction is too large to fit in one word for a reasonable wrodlength www.bookspar.com | Website for students | VTU NOTES

  22. Accomplish C = A + B using two-address instructions • Operation Source, Destination • Eg., Add A,B which performs the operation B  [A] + [B] • When sum is calculated result is sent to memory replacing original contents of location B • We cannot use a single two-address instruction as we don’t want to destroy the contents of A or B • Solution? • Use a instruction which copies contents of one memory location to another • Move B,C performs the operation C  [B] leaving contents of B unchanged • Actually it only “copies” not “moves” • Final solution is • Move B,C • Add A,C www.bookspar.com | Website for students | VTU NOTES

  23. Accomplish C = A + B using one-address instructions • Even two-address instructions will not fit into one single word of memory • Use one-address instruction • Second operand , whenever required, is present in a unique location • Eg., usage of accumulator, a processor register • Add A • means Add the contents of memory location A to the contents of accumulator register and place the sum back into the accumulator • Load A • Copies contents of memory location A into accumulator • Store A • Copies contents of accumulator to memory location A • Solution? www.bookspar.com | Website for students | VTU NOTES

  24. Contd… • Load A • Add B • Store C • Operand specified may be source or destination depending on the instruction • For load it is source, • For store it is destination www.bookspar.com | Website for students | VTU NOTES

  25. Accomplish C = A + B using one-address instructions, and General purpose registers • Most modern computers have 32 General Purpose Registers or more • Only 5 bits sufficient to address 32 GPRs. How? • Registers are used to store data temporarily • Since access time to registers is much less than memory frequent access to memory is reduced hence enhancing speed www.bookspar.com | Website for students | VTU NOTES

  26. Accomplish C = A + B using one-address instructions, and General purpose registers • If Ri is a GPR • Load A,Ri • Store Ri,A • Add A,Ri are generalizations for the single-accumulator case • Data transfer instruction - Move • When we want to move from one place to another. • A single instruction Move can be used in place of Load and Store • Move Source, Destination • Move A, Ri is same as Load A,Ri • Move Ri,A is same as Store Ri,A www.bookspar.com | Website for students | VTU NOTES

  27. Contd • If artihmetic operations are allowed only on operands in processor registers then to achieve C = A + B • Move A,Ri • Move B,Rj • Add Ri,Rj • Move Rj,C • If one operand can be in memory but other must be in register then • Move A,Ri • Add B,Rj • Move Rj,C www.bookspar.com | Website for students | VTU NOTES

  28. Zero-Address instructions • Locations of all operands defined implicitly • Machines which store operands in a structure called pushdown stack • Instructions does not specify any memory address www.bookspar.com | Website for students | VTU NOTES

  29. Instruction execution and straight line sequencing • Assumptions • GPRs • One addr insts • Wordlength 32bits • Byte addressable • Full address in • Single word instruction • The three insts at • i, i + 4, i + 8 Address Contents 3-instruction i Begin execution here Move A,R0 program i + 4 Add B,R0 segment i + 8 Move R0,C A Data for B the program C Figure 2.8. A program for C ¬ [A] + [B]. www.bookspar.com | Website for students | VTU NOTES

  30. Instruction execution and straight line sequencing • PC holds address of instruction to be executed next • To begin execution • place the address of first inst to be executed in PC • Processor control circuits use info in PC to fetch and execute instructions, one at a time, in order of increasing addresses • This is called as Straight-line sequencing • During the execution of each instruction, PC is incremented by 4 www.bookspar.com | Website for students | VTU NOTES

  31. Instruction execution and straight line sequencing • Execution is two phase • Instruction Fetch • Instruction is fetched from memory location whose address in PC • This instruction is placed in IR • Instruction Execute • IR is examined to determine which operation is to be performed • Specified operation is then performed by processor • Involves fetching operands from memory or processor registers, performing arithmetic or logic operations, storing results in destination • At some point during this two-phase, the contents of PC is advanced to point to next instruction www.bookspar.com | Website for students | VTU NOTES

  32. Branching • Branch instruction load a new value to PC than the address of next immediate instruction following the branch instruction • This new address is called branch target • A conditional branch instruction causes branch only if a specified condition is satisfied Eg., Branch>0 LOOP www.bookspar.com | Website for students | VTU NOTES

  33. Move NUM1,R0 i i + 4 Add NUM2,R0 i + 8 Add NUM3,R0 • • • i Add NUM n ,R0 + 4 n - 4 i 4 n + Move R0,SUM • • • SUM NUM1 NUM2 • • • NUM n Figure 2.9. A straight-line program for addingn numbers. www.bookspar.com | Website for students | VTU NOTES

  34. Move N,R1 Clear R0 LOOP Determine address of "Next" number and add "Next" number to R0 Program loop Decrement R1 Branch>0 LOOP Move R0,SUM • • • SUM N n NUM1 NUM2 • • • NUM n Figure 2.10. Using a loop to addn numbers. www.bookspar.com | Website for students | VTU NOTES

  35. Assignment to be submitted on Thursday • Express the following signed numbers in 2’s complement notation and perform addition and subtraction. State whether overflow occurs or not • 4-bit notation • 2&3, 5 & -6, -7 & 6, -8 & -3, 7 & 4 • 5-bit notation • 12 & 3, 7 & -7, -6 & 14, -10 & -4, 12 & 8 • Represent the following numbers in 32-bit Big-endian and Little-endian memory organization • 81234561 • -81234561 www.bookspar.com | Website for students | VTU NOTES

  36. Indirection and pointers • Indirect mode – the Effective address of the operand is the contents of a register or memory location whose address appears in the instruction • The register of memory location that contains the address of an operand is called a pointer • Analogy of treasure hunt • Instead of finding the treasure, we find the address where we find the treasure • By changing the contents of register R1 or location A in the following figure, the same add instruction fetches different operands to add to register R0. www.bookspar.com | Website for students | VTU NOTES

  37. Add (R1),R0 Add (A),R0 Main memory B Operand A B Register Operand R1 B B (a) Through a general-purpose register (b) Through a memory location Figure 2.11. Indirect addressing. www.bookspar.com | Website for students | VTU NOTES

  38. Indirect addressing logic for the program of adding n numbers using loop Address Contents Move N,R1 Initialization Move #NUM1,R2 Clear R0 LOOP Add (R2),R0 Add #4,R2 Decrement R1 LOOP Branch>0 Move R0,SUM Figure 2.12. Use of indirect addressing in the program of Figure 2.10. www.bookspar.com | Website for students | VTU NOTES

  39. INDEXED ADDRESSING Add 20(R1),R2 R1 1000 1000 20 = offset 1020 Operand (a) Offset is given as a constant Add 1000(R1),R2 1000 20 R1 20 = offset 1020 Operand (b) Offset is in the index register Figure 2.13. Indexed addressing. www.bookspar.com | Website for students | VTU NOTES

  40. n N Student ID LIST LIST + 4 Test 1 Student 1 LIST + 8 Test 2 LIST + 12 Test 3 LIST + 16 Student ID Test 1 Student 2 Test 2 Test 3 • • • Figure 2.14. A list of students' marks. www.bookspar.com | Website for students | VTU NOTES

  41. Note: contents of R0, which is used as indexed register, are not changed when it is used in indexed addressing mode to access test scores. Contents of R0 change only in last Add instruction, to move from one student record to the next Move #LIST,R0 Clear R1 Clear R2 Clear R3 Move N,R4 Add 4(R0),R1 LOOP Add 8(R0),R2 A d d 12(R0),R3 Add #16,R0 Decrement R4 Branch>0 LOOP Move R1,SUM1 R2,SUM2 Move Move R3,SUM3 Figure 2.15. Indexed addressing used in accessing test scores in the list in Figure 2.14. www.bookspar.com | Website for students | VTU NOTES

  42. Indexed addressing contd… • Several variations of the basic form of indexed addressing provide efficient access to memory operands • A second register may be used to contain the offset X, called as based indexed addressing mode • Denoted as ( Ri, Rj ) • Effective address is the sum of contents of Ri and Rj • Second register is called base register. • Provides more flexibility to the user • Eg., suppose, in the previous example, instead of only 3 items, each student record contain a large no of items, say k • We can replace the three Add instructions by a single instruction inside a second loop www.bookspar.com | Website for students | VTU NOTES

  43. Indexed addressing contd… • Problem • The list of student marks shown in prev figure 2.14, is changed to contain j test scores for each student. Assume that there are n students. Write an assembly language program for computing the sums of the scores on each test and store these sums in the memory word locations at addresses SUM, SUM + 4, SUM + 8,…. The number of tests, j, is larger than the number of registers in the processor. Use two nested loops, the inner loop should accumulate the sum for a particular test, and the outer loop should run over the number of tests, j. assume that j is stored in memory location J. www.bookspar.com | Website for students | VTU NOTES

  44. Solution www.bookspar.com | Website for students | VTU NOTES

  45. Indexed addressing mode contd.. • based indexed addressing mode with offset • Uses two registers plus a constant • X( Ri, Rj ) - Effective address is the sum of the constant X and the contents of registers Ri and Rj • This mode implements 3-Dimensional array www.bookspar.com | Website for students | VTU NOTES

  46. Relative addressing • Till now for index mode, registers we used are general purpose registers • A useful variation of this is to use Program Counter PC instead of a general purpose register • X(PC) – to address a memory location that is X bytes away from the location presently pointed to by the program counter • Relative mode- the effective address is determined by the index mode using the program counter in place of the general-purpose register Ri. • MOST common use is to specify the target address in branch instructions • Eg., Branch>0 LOOP • If branch condition is true, the program execution goes to branch target location identified by name LOOP • Can compute this location by specifying it as an offset from the current value of the program counter. • Since branch target address can be either before or after the branch instruction, the offset is given as a signed number www.bookspar.com | Website for students | VTU NOTES

  47. Example for relative addressing mode • Assume 4 instructions starting from LOOP are located at memory locations 1000,1004,1008 & 1012. • Updated contents at the time branch target address is generated is 1016. to branch to location LOOP(1000), the offset value needed is X = -16 Address Contents Move N,R1 Initialization Move #NUM1,R2 Clear R0 LOOP Add (R2),R0 Add #4,R2 Decrement R1 LOOP Branch>0 Move R0,SUM www.bookspar.com | Website for students | VTU NOTES

  48. Additional modes • Autoincrement and autodecrement mode – useful for accessing data items in successive locations in the memory • Autoincrement mode – • The effective address of the operand is the contents of a register specified in the instruction. After accessing the operand, the contents of this register are automatically incremented to point to next element in the list • (Ri)+ • The increment is 1 for byte-sized operands, 2 for 16-bit operands and 4 for 32-bit operands. Usually size of operand is specified in the instruction www.bookspar.com | Website for students | VTU NOTES

  49. Autoincrement addressing logic for the program of adding n numbers using loop Move N,R1 Initialization Move #NUM1,R2 Clear R0 LOOP Add (R2)+,R0 Decrement R1 Branch>0 LOOP Move R0,SUM Figure 2.16. The Autoincrement addressing mode used in the program of Figure 2.12. www.bookspar.com | Website for students | VTU NOTES

  50. Autodecrement mode • The contents of a register specified in the instruction are first automatically decremented and are then used as the effective address of the operand • Denoted by –(Ri) • Operands are accessed in descending address order. • The way the autoincrement and autodecrement modes are specified in very useful when implementing an important data structure called a stack. • Always we can use two instructions to perform the functionality of autoincrement and autodecrement mode www.bookspar.com | Website for students | VTU NOTES

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