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ECE 15B Computer Organization Spring 2010 Dmitri Strukov Lecture 4: Arithmetic / Data Transfer Instructions Partially adapted from Computer Organization.

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Presentation on theme: "ECE 15B Computer Organization Spring 2010 Dmitri Strukov Lecture 4: Arithmetic / Data Transfer Instructions Partially adapted from Computer Organization."— Presentation transcript:

1 ECE 15B Computer Organization Spring 2010 Dmitri Strukov Lecture 4: Arithmetic / Data Transfer Instructions Partially adapted from Computer Organization and Design, 4 th edition, Patterson and Hennessy, and classes taught by Ryan Kastner at UCSB

2 Agenda Review of last lecture Load/store operations Multiply and divide instructions ECE 15B Spring 2010

3 Last Lecture ECE 15B Spring 2010

4 Assembly Language ECE 15B Spring 2010 Basic job of a CPU: execute lots of instructions Instructions are the primitive operations that the CPU may execute Different CPUs implement different sets of instructions Instruction Set Architecture (ISA) is a set of instructions a particular CPU implements Examples: Intel 80x86 (Pentium 4), IBM/Motorola Power PC (Macintosh), MIPS, Intel IA64, ARM

5 Assembly Variables: Registers Unlike HLL like C or Java, assembly cannot use variables – Why not? Keep hardware simple Assembly Operands are registers – Limited number of special locations built directly into the hardware – Operations can only be performed on these – Benefit: Since registers file is small, it is very fast ECE 15B Spring 2010

6 Assembly Variables: Registers By convention, each register also has a name to make it easier to code For now: $16 - $23  $s0 - $s7 (correspond to C variables) $8 - $15  $t0 - $t7 (correspond to temporary variables) Will explain other 16 register names later In general, use names to make your code more readable ECE 15B Spring 2010

7 MIPS Syntax Instruction Syntax: [Label:] Op-code [oper. 1], [oper. 2], [oper.3], [#comment] (0) (1) (2) (3) (4) (5) – Where 1) operation name 2,3,4) operands 5) comments 0) label field is optional, will discuss later – For arithmetic and logic instruction 2) operand getting result (“destination”) 3) 1 st operand for operation (“source 1”) 4) 2 nd operand for operation (source 2” Syntax is rigid – 1 operator, 3 operands – Why? Keep hardware simple via regularity ECE 15B Spring 2010

8 Addition and Subtraction of Integers Addition in assembly – Example: add $s0, $s1, $s2 (in MIPS) Equivalent to: a = b + c (in C) Where MIPS registers $s0, $s1, $s2 are associated with C variables a, b, c Subtraction in Assembly – Example Sub $s3, $s4, S5(in MIPS) Equivalent to: d = e - f (in C) Where MIPS registers $s3, $s4, $s5 are associated with C variables d, e, f ECE 15B Spring 2010

9 Addition and Subtraction of Integers How do we do this? f = (g + h) – (i + j) Use intermediate temporary registers add $t0, $s1, $s2#temp = g + h add $t1, $s3, $s4#temp = I + j sub $s0, $t0, $t1#f = (g+h)-(i+j) ECE 15B Spring 2010

10 Immediates Immediates are numerical constants They appear often in code, so there are special instructions for them Add immediate: addi $s0, $s1, 10# f= g + 10 (in C) – Where MIPS registers $s0 and $s1 are associated with C variables f and g – Syntax similar to add instruction, except that last argument is a number instead of register ECE 15B Spring 2010

11 Load and Store Instructions ECE 15B Spring 2010

12 CPU Overview ECE 15B Spring 2010

13 … with muxes Can’t just join wires together Use multiplexers ECE 15B Spring 2010

14 … with muxes ECE 15B Spring 2010

15 Memory Operands Main memory used for composite data – Arrays, structures, dynamic data To apply arithmetic operations – Load values from memory into registers – Store result from register to memory Memory is byte addressed – Each address identifies an 8-bit byte Words are aligned in memory – Address must be a multiple of 4 MIPS is Big Endian – Most-significant byte at least address of a word – c.f. Little Endian: least-significant byte at least address ECE 15B Spring 2010

16 Data Transfer: Memory to Register MIPS load Instruction Syntax lw register#, offset(register#) (1) (2) (3) (4) Where 1) operation name 2) register that will receive value 3) numerical offset in bytes 4) register containing pointer to memory lw – meaning Load Word 32 bits or one word are loaded at a time ECE 15B Spring 2010

17 Data Transfer: Register to Memory MIPS store Instruction Syntax sw register#, offset(register#) (1) (2) (3) (4) Where 1) operation name 2) register that will be written in memory 3) numerical offset in bytes 4) register containing pointer to memory sw – meaning Store Word 32 bits or one word are stored at a time ECE 15B Spring 2010

18 Memory Operand Example 1 C code: g = h + A[8]; – g in $s1, h in $s2, base address of A in $s3 Compiled MIPS code: – Index 8 requires offset of 32 4 bytes per word lw $t0, 32($s3) # load word add $s1, $s2, $t0 offset base register ECE 15B Spring 2010

19 Memory Operand Example 2 C code: A[12] = h + A[8]; – h in $s2, base address of A in $s3 Compiled MIPS code: – Index 8 requires offset of 32 lw $t0, 32($s3) # load word add $t0, $s2, $t0 sw $t0, 48($s3) # store word ECE 15B Spring 2010

20 Registers vs. Memory Registers are faster to access than memory Operating on memory data requires loads and stores – More instructions to be executed Compiler must use registers for variables as much as possible – Only spill to memory for less frequently used variables – Register optimization is important! ECE 15B Spring 2010

21 Byte/Halfword Operations MIPS byte/halfword load/store – String processing is a common case lb rt, offset(rs) lh rt, offset(rs) – Sign extend to 32 bits in rt lbu rt, offset(rs) lhu rt, offset(rs) – Zero extend to 32 bits in rt sb rt, offset(rs) sh rt, offset(rs) – Store just rightmost byte/halfword Why do we need them?  characters and multimedia data are expressed by less than 32 bits; having dedicated 8 and 16 bits load and store instructions results in faster operation ECE 15B Spring 2010

22 Two’s Compliment Representation Multiply and Divide ECE 15B Spring 2010

23 Unsigned Binary Integers Given an n-bit number Range: 0 to +2 n – 1 Example 0000 0000 0000 0000 0000 0000 0000 1011 2 = 0 + … + 1×2 3 + 0×2 2 +1×2 1 +1×2 0 = 0 + … + 8 + 0 + 2 + 1 = 11 10 Using 32 bits 0 to +4,294,967,295 ECE 15B Spring 2010

24 2s-Complement Signed Integers Given an n-bit number Range: –2 n – 1 to +2 n – 1 – 1 Example 1111 1111 1111 1111 1111 1111 1111 1100 2 = –1×2 31 + 1×2 30 + … + 1×2 2 +0×2 1 +0×2 0 = –2,147,483,648 + 2,147,483,644 = –4 10 Using 32 bits –2,147,483,648 to +2,147,483,647 ECE 15B Spring 2010

25 2s-Complement Signed Integers Bit 31 is sign bit – 1 for negative numbers – 0 for non-negative numbers –(–2 n – 1 ) can’t be represented Non-negative numbers have the same unsigned and 2s-complement representation Some specific numbers – 0:0000 0000 … 0000 – –1:1111 1111 … 1111 – Most-negative:1000 0000 … 0000 – Most-positive:0111 1111 … 1111 ECE 15B Spring 2010

26 Signed Negation Complement and add 1 – Complement means 1 → 0, 0 → 1 Example: negate +2 +2 = 0000 0000 … 0010 2 –2 = 1111 1111 … 1101 2 + 1 = 1111 1111 … 1110 2 ECE 15B Spring 2010

27 Sign Extension Representing a number using more bits – Preserve the numeric value In MIPS instruction set – addi : extend immediate value – lb, lh : extend loaded byte/halfword – beq, bne : extend the displacement Replicate the sign bit to the left – c.f. unsigned values: extend with 0s Examples: 8-bit to 16-bit – +2: 0000 0010 => 0000 0000 0000 0010 – –2: 1111 1110 => 1111 1111 1111 1110 ECE 15B Spring 2010

28 Integer Addition Example: 7 + 6 ECE 15B Spring 2010

29 Integer Subtraction Add negation of second operand Example: 7 – 6 = 7 + (–6) +7:0000 0000 … 0000 0111 –6:1111 1111 … 1111 1010 +1:0000 0000 … 0000 0001 ECE 15B Spring 2010

30 Multiplication Start with long-multiplication approach 1000 × 1001 1000 0000 1000 1001000 Length of product is the sum of operand lengths multiplicand multiplier product ECE 15B Spring 2010

31 Multiplication Hardware Initially 0 ECE 15B Spring 2010

32 Stopped here… will start next lecture from here ECE 15B Spring 2010

33 Optimized Multiplier Perform steps in parallel: add/shift One cycle per partial-product addition That’s ok, if frequency of multiplications is low ECE 15B Spring 2010

34 Faster Multiplier Uses multiple adders – Cost/performance tradeoff Can be pipelined Several multiplication performed in parallel ECE 15B Spring 2010

35 MIPS Multiplication Two 32-bit registers for product – HI: most-significant 32 bits – LO: least-significant 32-bits Instructions – mult rs, rt / multu rs, rt 64-bit product in HI/LO – mfhi rd / mflo rd Move from HI/LO to rd Can test HI value to see if product overflows 32 bits – mul rd, rs, rt Least-significant 32 bits of product –> rd ECE 15B Spring 2010

36 Division Check for 0 divisor Long division approach – If divisor ≤ dividend bits 1 bit in quotient, subtract – Otherwise 0 bit in quotient, bring down next dividend bit Restoring division – Do the subtract, and if remainder goes < 0, add divisor back Signed division – Divide using absolute values – Adjust sign of quotient and remainder as required 1001 1000 1001010 -1000 10 101 1010 -1000 10 n-bit operands yield n-bit quotient and remainder quotient dividend remainder divisor ECE 15B Spring 2010

37 Division Hardware Initially dividend Initially divisor in left half ECE 15B Spring 2010

38 Optimized Divider One cycle per partial-remainder subtraction Looks a lot like a multiplier! – Same hardware can be used for both ECE 15B Spring 2010

39 Faster Division Can’t use parallel hardware as in multiplier – Subtraction is conditional on sign of remainder Faster dividers (e.g. SRT devision) generate multiple quotient bits per step – Still require multiple steps ECE 15B Spring 2010

40 MIPS Division Use HI/LO registers for result – HI: 32-bit remainder – LO: 32-bit quotient Instructions – div rs, rt / divu rs, rt – No overflow or divide-by-0 checking Software must perform checks if required – Use mfhi, mflo to access result ECE 15B Spring 2010

41 Conclusions In MIPS assembly language – Register replace C variables – One instruction (simple operation) per line – Simpler is faster Memory is byte-addressable, but lw and sw access one word at a time A pointer (used by lw and sw) is just a memory address, so we can add to it or subtract from it (using offset) ECE 15B Spring 2010

42 Review Instructions so far: add, addi, sub mult, div, mfhi, mflo, lw, sw, lb, lbu, lh, lhu Registers so far C variables: $s0 - $s7 Temporary variables: $t0 - $t9 Zero: $zero ECE 15B Spring 2010


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