COMPUTER ORGANIZATION NOTES Unit 2

COMPUTER ORGANIZATION NOTES Unit 2

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  Unit II: MACHINEINSTRUCTIONS and PROGRAMS This unit deals with concepts as to how programs are executed in the computer from the machine instruction set viewpoint. All computers deal with numbers. They have instructions that perform operations on the data operands. Hence we start with basic Number systems. NUMBER, ARITHMETIC OPERATIONS and CHARACTERS Binary numbers (0, 1) are used in computers. Various number systems are used in computers. Numbers in binary are represented as vectors B=bn-1…b1b0 Unsigned numbers are in range 0 to 2n-1 and are represented by V (B) =bn-1×2n-1 +…+b1 ×21 +b0 ×20 MSB LSB Negative Numbers: They can be represented in various ways given below. • Sign-and-magnitude Most significant bit determines sign, remaining unsigned bits represent magnitude • 1’s complement Most significant bit determines sign. To change sign from unsigned to negative, invert all the bits ( -3 is obtained by complementing each bit in vector 0011 to yield 1100). • 2’s complement Most significant bit determines sign. To change sign from unsigned to negative, invert all the bits and add 1. This is equivalent to subtracting the positive number from 2n. The representations are as given in the table below
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  Figure 1: Binary,signed integer representation Addition & Subtraction of Signed Numbers: 3 systems of representing signed numbers These systems differ only in the way they represent negative number Sign and magnitude system – simplest representation – most awkward for addition and subtraction. 1’s complement method is somewhat better. 2’s complement is the most efficient method Circle representation of Integer Mod N: This is a graphical technique to compute (a+b) mod 16. This can be also used for addition involving signed numbers. Both the cases are shown below Figure 2:Circle representation of integer Mod 16
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  The operation (7+4)mod 16 yields the value 11. To perform this graphically using the above representation locate 7 on the circle and then move 4 units in the clock wise direction to arrive at the answer 11. Next let us consider adding +7 to -3. The representation is as shown below Figure 3: Mod 16 system for 2’s complement numbers 2’s complement representation for 7 is 0111 and -3 is 1101. Thus locate 0111 and then move 1101(13 steps) in clockwise direction to arrive at 0100 = +4. Some more examples of 2’s complement add and substract operations are as shown in figure below: Figure 4: 2’s complement add and substract operations
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  Overflow in integerarithematic: In 2’s complement arithmetic addition of opposite sign numbers will never result in overflow . If the numbers are the same sign and the result is the opposite sign, overflow has occurred. E.g. 0100 + 0111 =1011 (but 1011 is -5) In case of unsigned numbers carry out signals that an overflow has occured Characters: Apart from numbers computers must be able to handle alphanumeric text information consisting of characters . Characters can be letters of alphabets, decimal digits , punctuation marks etc. Most widely used code was ASCII and now a days unicode is being used widely. Figure 5: ASCII Codes Figure 6: Example of Uni Code
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  MEMORY LOCATIONS andADDRESSES Memory consists of storage cells. They store the bits 0 or 1. We can deal with them in n-bit groups called words (typically 8, 16, 32 or 64 bits). Usually refer to memory size in bytes e.g. we say we have 128MB memory and rarely use words as the unit. We use addresses to store or retrieve item of information For some k, memory consists of 2k unique addresses which range from 0 to 2k -1. The possible addresses are the address space of the computer. E.g. 24-bit address has 2 24 (16,777,216) locations. Information quantities: bit, byte, word where Byte=8 bits, word typically varies 16-64 bits. Most machines address memory in units of bytes. For a 32-bit machine, successive words are at address 0, 4, 8, 12 and so on. Significant Bytes: Consider the hexadecimal (base 16) 32-bit number 34123F2A. This number is made up of four bytes 34, 12, 3F, 2A (4x8=32-bits). Bytes/bits with higher weighting are “more significant” i.e. the byte 34 is more significant than 2A. Bytes/bits with lower weighting are “less significant” i.e. 2A. Two ways byte addresses can be assigned across words – More significant bytes first (big endian) - SUN/SPARC, IBM/RISC6000 – Less significant bytes first (little endian) - Intel Pentium Processors Figure 7: Byte and Word addressing
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  Big Endian andLittle Endian: Consider a 32 bit integer (in hex): 0xabcdef12. It consists of 4 bytes: ab, cd, ef, and 12. Hence this integer will occupy 4 bytes in memory. Say we store it at memory address starting 1000. There are 24 different orderings possible to store these 4 bytes in 4 locations (1000 - 1003). 2 among these 24 possibilities are very popular. These are called as little endian and big endian. INSTRUCTIONS and INSTRUCTION SEQUENCING A computer must have instruction capable of performing the following operations. They are: • Data transfer between memory and processor register. • Arithmetic and logical operations on data. • Program sequencing and control. • I/O transfer. Register Transfer Notation: The possible locations that may be involved during data transfer are Memory Location Processor register Registers in I/O sub-system. Location Hardware Binary Address Expression Description Memory LOC,PLACE,A,VAR2 R1 [LOC] The contents of memory location LOC are transferred to the processor register R1. Processor R0,R1, R2…Rn [R3] [R1]+[R2] Add the contents of register R1 &R2 and the result of the operation is stored into register R3.
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  Assembly Language Notation: AssemblyLanguage Format Description Move LOC,R1 Transfers the contents of memory location LOC to the processor register R1. Add R1,R2,R3 Add the contents of register R1 & R2 and stores their sum into register R3. Basic Instruction Type: Instruction Type Syntax Expression Description Three Address Operation Source1,Source2,Destination Add A,B,C C←[A]+[B] Two Address Operation Source,Destination Add A,B B←[A]+[B] One Address Operation Operand Add B Content of B added with the content of the accumulator INSTRUCTION EXECUTION and STRAIGHT LINE SEQUENCING Instruction Execution: There are 2 phases for executing an instruction. They are, • Instruction Fetch • Instruction Execution Instruction Fetch: The instruction is fetched from the memory location whose address is in PC. This is then placed in IR. Instruction Execution: Instruction in IR is examined and decoded to determine which operation is to be performed. Program execution Steps: To begin executing a program, the address of first instruction must be placed in PC. The processor control circuits use the information in the PC to fetch & execute instructions one at a time in the order of increasing order. This is called Straight line sequencing. During the execution of each instruction, the PC is incremented by 4 to point to the address of next instruction.
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  Figure 8:A programfor C [A]+ [B] Branching: The Address of the memory locations containing the n numbers are symbolically given as NUM1, NUM2…..NUMn. Separate Add instruction is used to add each number to the contents of register R0. After all the numbers have been added, the result is placed in memory location SUM. Figure 9:A straight line program for adding n numbers
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  Using loop toadd ‘n’ numbers: • Number of entries in the list „n is stored in memory location M. Register R1 is used as a counter to determine the number of times the loop is executed. • Content location M are loaded into register R1 at the beginning of the program. • It starts at location Loop and ends at the instruction, Branch>0.During each pass, the address of the next list entry is determined and the entry is fetched and added to R0. • Decrement R1; It reduces the contents of R1 by 1 each time through the loop. • Branch >0 Loop; A conditional branch instruction causes a branch only if a specified condition is satisfied. Figure 10:Using a loop to add n numbers Conditional Codes:In order to do conditional branches and other instructions, operations implicitly set flags. Four commonly used (1-bit) flags • N (negative) 1 if result –ve else 0 • Z (zero) 1 if result 0 else 0 • V (overflow) 1 if arithmetic overflow occurs else 0 • C (carry) 1 if carry out occurs –ve else 0
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  ADDRESSING MODE The differentways in which the location of an operand is specified in an instruction is called as Addressing mode. Generic Addressing Modes: • Immediate mode • Register mode • Absolute mode • Indirect mode • Index mode • Base with index • Base with index and offset • Relative mode • Auto-increment mode • Auto-decrement mode Implementation of Variables and Constants: Variables: The value can be changed as needed using the appropriate instructions. There are 2 accessing modes to access the variables. They are • Register Mode • Absolute Mode Register Mode: The operand is the contents of the processor register. The name (address) of the register is given in the instruction. Absolute Mode (Direct Mode): The operand is in a memory location. The address of this location is given explicitly in the instruction. The various addressing modes and their assembler syntax and functions are as shown in figure below: Figure 11:Generic addressing modes
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  Immediate Addressing Mode:The operand is specified in the instruction itself . Move 200immediate, R0 Move #200, R0 Direct Addressing Mode: Operand resides in Memory and its address is given explicitly in the address field of an instruction. Move P, R0 Move R0, S Add Q, R0 Register Addressing Mode: name of the register (address code of a specific general purpose register) appears in the address field of an instruction i.e. Move B, R1 Indexing and Arrays: Index Mode: The effective address of an operand is generated by adding a constant value to the contents of a register. The constant value uses either special purpose or general purpose register. X (RI) where X – denotes the constant value contained in the instruction Ri – name of the register involved. The Effective Address of the operand EA=X + [Ri] The index register R1 contains the address of a new location and the value of X defines an offset (also called a displacement). To find operand first go to Reg R1 (using address)-read the content from R1 i.e. 1000 Add the content 1000 with offset 20 to get the result. Here the constant X refers to the new address and the contents of index register that defines the offset to the operand. The sum of two values is given explicitly in the instruction and the other is stored in register. Add 20(R1), R2 (or) EA=>1000+20=1020 Relative Addressing: It is same as index mode. The difference is, instead of general purpose register, here we can use program counter (PC). Relative Mode: The Effective Address is determined by the Index mode using the PC in place of the general purpose register. This mode can be used to access the data operand. But it’s most common use is to specify the target address in branch instruction. Eg. Branch>0 Loop It causes the program execution to go to the branch target location. It is identified by the name loop if the branch condition is satisfied.
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  Additional Modes: There aretwo additional modes. They are • Auto-increment mode • Auto-decrement mode Auto-increment mode: The Effective Address of the operand is the contents of a register in the instruction. After accessing the operand, the contents of this register is automatically incremented to point to the next item in the list. Auto-decrement mode: The Effective Address of the operand is the contents of a register in the instruction. After accessing the operand, the contents of this register is automatically decremented to point to the next item in the list. Assembly Language We generally use symbolic names to write a program. A complete set of such symbolic names and rules for their use constitute a programming language, is referred to as assembly language. LOAD - To load operand from memory STORE - To store operand to memory MOVE - To transfer data from one location to another location/Register Assembler Directives: Directives are the assembler commands to the assembler concerning the program being assembled. These commands are neither translated into machine opcode nor assigned any memory location in the object program. S EQU 150 EQU directs the assembler that the symbolic name S must be replaced with memory location address 150, ORIGIN 201 Instruct assembler to initiate data block at main memory locations starting from 201 N DATAWORD 40 Inform the assembler that value of N i.e. data value 40 is to be placed in the memory location 201. ORIGIN 100 States that assembler directive must load machine instructions of the object program in the main memory starting from location 100. END START End of the program and the label of where program starts N1 RESERVE 400 Reserve memory block of 400 bytes
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  Assembler: Has toknow • How to interpret machine language (directives, instructions, addressing modes etc) • Where to place the instructions in memory • Where to place the data in memory ; Scans through source program, keeps track of all names and corresponding numerical values in symbol table e.g. what all the labels mean • Calculate branch addresses; Forward branch problem – how can it work out forward addresses? Two Pass Assembler: • First pass – Work out all the addresses of labels • Second pass – Generate machine code, substituting values for the labels BASIC INPUT/OUTPUT OPERATIONS I/O is the means by which data are transferred between the processor and the outside world. Devices operate at different speeds to the processor so handshaking is required. Keyboard/display Example: The keyboard and display are coordinated via software • Register (on device) assigned to the keyboard hardware – DATAIN contains ASCII of last typed character – SINis the status control flag, normally 0. When a character typed, becomes 1. After the processor reads DATAIN, it is automatically set back to 0 • Register (on device) assigned to the display hardware – DATAOUT receives a character code – SOUTis the status control flag. It is 1 when ready to receive a character, set to 0 when the character is being transferred • These registers form the respective device interface Figure 12: Bus Connection for processor, keyboard and display
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  Memory mapped IOand IO mapped IO: Figure 13: Memory mapped IO and IO mapped IO I/O Driver program (Programmed IO): READWAIT Branch to READWAIT if SIN=0 INPUT from DATAIN to R1 WRITEWAIT Branch to WRITEWAIT if SOUT=0 Output from R1 to DATAOUT Memory Mapped IO: • On many machines, registers such as DATAIN, DATAOUT are memory-mapped – Read and write specific memory locations to communicate with device – Move Byte DATAIN, R1 – Move Byte R1,DATAOUT • SIN and SOUT might be bits in a device status register e.g. bit 3 READWAIT Branch to READWAIT if SIN=0 INPUT from DATAIN to R1 READWAIT Test bit #3, INSTATUS Branch=0 READWAIT Move Byte DATAIN, R1
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  WRITEWAIT Branch toWRITEWAIT if SOUT=0 Output from R1 to DATAOUT WRITEWAIT Test bit #3, OUTSTATUS Branch=0 WRITEWAIT Move Byte R1, DATAOUT Program to read a line of character and display it : Move #LOC,R0 Initialize pointer register R0 to point to the address of the first location in memory where the characters are to be stored. READ TestBit #3,INSTATUS Wait for a character to be entered Branch=0 READ in the keyboard buffer DATAIN. MoveByte DATAIN,(R0) Transfer the character from DATAIN into the memory (this clears SIN to 0) ECHO TestBit #3, OUTSTATUS Wait for the display to become ready. Branch=0 ECHO MoveByte (R0),DATAOUT Move the character just read to the display buffer register (this clears SOUT to 0). Compare #CR,(R0)+ Check if the character just read is CR (carriage return). If it is not CR, then Branch≠0 READ branch back and read another character. Also, increment the pointer to store the next character. STACKS and QUEUES List of data elements (usually bytes or words). Elements can only be removed at one end of the list. Last-in-first-out. Can be implemented in several ways, one way is – First element placed in BOTTOM – Grows in direction of decreasing memory address – Assume 32-bit data
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  Figure 14: Astack of words in the memory Stack Implementation: Subtract #4, SP Move NEWITEM,(SP) ; push Move (SP), ITEM; pop Add #4, SP With auto increment and auto decrement Move NEWITEM,-(SP); push Move (SP) +, ITEM; pop Queue: • First-in-first-out • Unlike a stack, need to keep track of both the front and end for removal and insertion respectively • Need two pointers to keep track of both ends • Assuming it moves through memory in direction of higher addresses, as it is used, it walks through memory towards higher addresses. Circular buffers: • Avoid this problem by limiting to a fixed region in memory • Start at BEGINNING and entries appended until it reaches END after which it wraps back around to BEGINNING • Need to deal with cases when it is completely full and completely empty
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  SUBROUTINES • Often needto perform subtask on different data we use subtask called a subroutine • Rather than include the same sequence of instructions everywhere it is needed, call a subroutine instead – One copy of subroutine stored in memory – Subroutine call causes a branch to the subroutine – At the end of the subroutine, a return instruction is executed – Program resumes execution at the instruction immediately following the subroutine call Parameter Passing: Subroutine call – e.g. SUM = listadd (N, NUM); – N is a variable in memory and NUM is an address pointing to the start of the NUM list – How do we send the parameters N, NUM to the subroutine? – How do we receive the return value SUM? One way is putting the parameters in registers; second way is Passing Parameters on stack and third way is passing by value and reference. CISC & RISC • Multiple length instructions are difficult to implement with high clock rate • Complex instruction set computers (CISC) have complex instruction encodings like this (e.g. IA-32) • Reduced instruction set computers (RISC) only allow simple 32-bit formats, few addressing modes and all data to be manipulated must be in registers e.g. Add (R3),R2 is not allowed, instead use Move (R3),R1 followed by Add R1,R2 (e.g. ARM) RISC machines often are 3-address machines as the addressing mode field is either not necessary or simplified e.g. Add R1, R2, R3 • CISC machines usually require less instructions but have a lower clock rate, RISC require more instructions but have a higher clock rate