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...