Chapter 3.4, Problem 3.5PP

Explanation of Solution

Given program:

xp in %rdi, yp in %rsi, zp in %rdx

decode1:

movq (%rdi), %r8  Get x = *xp

movq (%rsi), %rcx  Get y = *yp

movq (%rdx), %rax  Get z = *zp

movq  %r8, (%rsi)  Store x at yp

movq  %rcx, (%rdx)  Store y at zp

movq  %rax, (%rdi)  Store z at xp

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.

Example:

The example for “Mov” class instruction is shown below:

movq $-1, %rax

Here, “-1” represents “FF…F”, where count of “F” is twice count of bytes in representation. The given code sets value of “%rax” to “FFFFFFFFFFFFFFFF”...