CS170 Computer Organization and Architecture I

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CS170 Computer Organization and Architecture I
Ayman Abdel-Hamid Department of Computer
Science Old Dominion University Lecture 16
10/29/2002
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Outline

• Constant or Immediate Operands
• Pseudoinstrcutions
• Memory usage and a related pseudoinstruction

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Constant Operands1/6

• In the C compiler gcc, 52 of arithmetic
  operations involve constants. In circuit
  simulation program spice it is 69
• x x 4
• where x is associated with register s0, the
  constant 4 is stored in memory at memory address
  ADDR_4 What is the MIPS assembly code?

lw t0, ADDR_4(zero) t0 ? the constant 4 add
s0,s0,t0 s0 ? s0 4

• This solution requires memory access to load a
  constant from memory (not efficient)
• Offer versions of arithmetic instructions in
  which operand can be a constant
• Use I-format (Immediate-format) for such
  instructions (2 registers and a constant)
• The MIPS field containing the constant is 16 bits
  long

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Constant Operands2/6
Add Immediate instruction addi s0,s0,4 s0 ?
s0 4
Immediate version of slt slti t0,s2,10 t0
1 if s2 lt 10
Design principle Make the common case
fast Constant operands occur frequently, make
constants part of arithmetic instructions. Much
faster than a memory load
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Constant Operands3/6

• How about constants bigger than 16 bits?
• Set upper 16 bits of a constant in a register
  using lui instruction
• A subsequent instruction can specify the lower 16
  bits of the constant
• lui t0, 255 255 is 0000 0000 1111 1111
• Transfer 16 bit immediate constant into leftmost
  16 bits of register and fill lower 16 bits with
  0s.

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Constant Operands4/6
What is the MIPS assembly code to load the
following 32-bit constant into register s0? 0000
0000 0011 1101 0000 1001 0000 0000

• Load upper 16 bits using lui into s0
• lui s0, 61 6110 0000 0000 0011 11012
• Add lower 16 bits whose lower value is 2304
• addi s0, s0, 2304 230410 0000 1001 0000
  00002

• Compiler or assembler break large constants into
  pieces and then reassemble into a register
• If assembler performs such job, it needs a
  temporary register available in which to create
  the long values
• Register at is reserved for assembler

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Constant Operands5/6

• A word of caution
• addi sign extends 16-bit immediate field before
  performing addition
• Copy leftmost bit (sign bit) of 16-bit immediate
  field of instruction into upper 16 bits of a word
  before performing addition
• An alternative to addi is to use ori instruction
  (logical OR immediate) to create 32-bit constants
  in conjunction with lui (especially useful for
  base memory addresses)
• ori s0, s0,2304
• ori loads 0s into the upper 16 bits (zero-extend
  immediate field)
• Why it is OK to use ori after lui?

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Constant Operands6/6

• Any constant value can be expressed in decimal,
  binary, or hexadecimal
• By default the value is in decimal in
  instructions
• To represent in hex use 0x before constant value
• lui s0, 61 6110 0000 0000 0011 11012
• To represent constant in hex
• lui s0,0x003D Why 6110 is 003D16?

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Pseudoinstructions

• Some instructions are provided to simplify
  assembly language programming and translation,
  and give a richer set of instructions than those
  implemented by the hardware
• The assembler reads such instructions and
  replaces by a sequence of real machine
  instructions that have the same effect
• Example
• move t0,t1 pseudoinstruction t0 ? t1
• Assembler converts this instruction into machine
  language equivalent
• add t0, zero,t1 t0 ? 0 t1
• Other pseudoinstructions blt, bgt, bge, and ble
• You can use pseudoinstructions to simplify
  programming, but attempt to write your programs
  in terms of REAL MIPS instructions

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Memory Usage in MIPS

• See Figure A.9 (section A.5 in appendix A)
• Reserved for Operating System
• Program code (Text) starts at 0040 000016 (PC
  is program counter)
• Static data (data size known at compile time, and
  lifetime is the programs entire execution)
  starts at 1000 000016
• Dynamic data (allocation occurs during execution)
  is next and grows up towards the stack
• Stack is dynamic (will see function when we look
  at procedure calls)
• The register gp is the global pointer and is set
  to an address (1000 800016) to make it easy to
  access data

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A related pseudoinstruction

• How does an array address get loaded into a
  register?
• Array is allocated in the static data portion of
  the data segment
• We can use load address pseudoinstruction la
  rdest, symbolic address
• la s3, Address
• At compile time Address is known.
• Assume Address is 100000A416
• We need to put this value in s3 using real
  instructions
• Immediate field is only 16 bits (4 hex digits)?
• In this case, la gets implemented using two
  instructions
• lui s3, 0x1000
• addi s3,s3,0x00A4