EBK COMPUTER SYSTEMS
EBK COMPUTER SYSTEMS
3rd Edition
ISBN: 8220101459107
Author: O'HALLARON
Publisher: YUZU
Expert Solution & Answer
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Chapter 3.6, Problem 3.28PP

a.

Explanation of Solution

Given assembly code:

x in %rdi

fun_b:

movl $64, %edx

movl $0, %eax

.L10:

movq %rdi, %rcx

andl $1, %ecx

addq %rax, %rax

orq %rcx, %rax

shrq %rdi

subq $1, %rdx

jne .L10

rep; ret

Data movement instructions:

  • The different instructions are been grouped as “instruction classes”.
  • The instructions in a class performs same operation but with different sizes of operand.
  • The “Mov” class denotes data movement instructions that copy data from a source location to a destination.
  • The class has 4 instructions that includes:
    • movb:
      • It copies data from a source location to a destination.
      • It denotes an instruction that operates on 1 byte data size.
    • movw: 
      • It copies data from a source location to a destination.
      • It denotes an instruction that operates on 2 bytes data size.
    • movl:
      • It copies data from a source location to a destination.
      • It denotes an instruction that operates on 4 bytes data size.
    • movq:
      • It copies data from a source location to a destination.
      • It denotes an instruction that operates on 8 bytes data size.

Unary and Binary Operations:

  • The details of unary operations includes:
    • The single operand functions as both source as well as destination.
    • It can either be a memory location or a register.
    • The instruction “incq” causes 8 byte element on stack top to be incremented.
    • The instruction “decq” causes 8 byte element on stack top to be decremented.
  • The details of binary operations includes:
    • The first operand denotes the source.
    • The second operand works as both source as well as destination.
    • The first operand can either be an immediate value, memory location or register.
    • The second operand can either be a register or a memory location.

Jump Instruction:

  • The “jump” instruction causes execution to switch to an entirely new position in program.
  • The “label” indicates jump destinations in assembly code.
  • The “je” instruction denotes “jump if equal” or “jump if zero”.
    • The comparison operation is performed...

b.

Explanation of Solution

Initial test requirements:

  • The transformation guarded-do generate the code.
  • The compiler checks value of “i...

c.

Explanation of Solution

Explanation of function:

  • The bits in “x” are reversed using the code.
  • It creates a mirror image pattern of original bits...

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Chapter 3 Solutions

EBK COMPUTER SYSTEMS

Ch. 3.5 - Prob. 3.11PPCh. 3.5 - Prob. 3.12PPCh. 3.6 - Prob. 3.13PPCh. 3.6 - Prob. 3.14PPCh. 3.6 - Prob. 3.15PPCh. 3.6 - Prob. 3.16PPCh. 3.6 - Practice Problem 3.17 (solution page 331) An...Ch. 3.6 - Practice Problem 3.18 (solution page 332) Starting...Ch. 3.6 - Prob. 3.19PPCh. 3.6 - Prob. 3.20PPCh. 3.6 - Prob. 3.21PPCh. 3.6 - Prob. 3.22PPCh. 3.6 - Prob. 3.23PPCh. 3.6 - Practice Problem 3.24 (solution page 335) For C...Ch. 3.6 - Prob. 3.25PPCh. 3.6 - Prob. 3.26PPCh. 3.6 - Practice Problem 3.27 (solution page 336) Write...Ch. 3.6 - Prob. 3.28PPCh. 3.6 - Prob. 3.29PPCh. 3.6 - Practice Problem 3.30 (solution page 338) In the C...Ch. 3.6 - Prob. 3.31PPCh. 3.7 - Prob. 3.32PPCh. 3.7 - Prob. 3.33PPCh. 3.7 - Prob. 3.34PPCh. 3.7 - Prob. 3.35PPCh. 3.8 - Prob. 3.36PPCh. 3.8 - Prob. 3.37PPCh. 3.8 - Prob. 3.38PPCh. 3.8 - Prob. 3.39PPCh. 3.8 - Prob. 3.40PPCh. 3.9 - Prob. 3.41PPCh. 3.9 - Prob. 3.42PPCh. 3.9 - Practice Problem 3.43 (solution page 344) Suppose...Ch. 3.9 - Prob. 3.44PPCh. 3.9 - Prob. 3.45PPCh. 3.10 - Prob. 3.46PPCh. 3.10 - Prob. 3.47PPCh. 3.10 - Prob. 3.48PPCh. 3.10 - Prob. 3.49PPCh. 3.11 - Practice Problem 3.50 (solution page 347) For the...Ch. 3.11 - Prob. 3.51PPCh. 3.11 - Prob. 3.52PPCh. 3.11 - Practice Problem 3.52 (solution page 348) For the...Ch. 3.11 - Practice Problem 3.54 (solution page 349) Function...Ch. 3.11 - Prob. 3.55PPCh. 3.11 - Prob. 3.56PPCh. 3.11 - Practice Problem 3.57 (solution page 350) Function...Ch. 3 - For a function with prototype long decoda2(long x,...Ch. 3 - The following code computes the 128-bit product of...Ch. 3 - Prob. 3.60HWCh. 3 - In Section 3.6.6, we examined the following code...Ch. 3 - The code that follows shows an example of...Ch. 3 - This problem will give you a chance to reverb...Ch. 3 - Consider the following source code, where R, S,...Ch. 3 - The following code transposes the elements of an M...Ch. 3 - Prob. 3.66HWCh. 3 - For this exercise, we will examine the code...Ch. 3 - Prob. 3.68HWCh. 3 - Prob. 3.69HWCh. 3 - Consider the following union declaration: This...Ch. 3 - Prob. 3.71HWCh. 3 - Prob. 3.72HWCh. 3 - Prob. 3.73HWCh. 3 - Prob. 3.74HWCh. 3 - Prob. 3.75HW
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