Chapter 3, Problem 3.72HW

A.

Explanation of Solution

Given Assembly code:

# long aframe(long n, long idx, long *q)

# n in %rdi, idx in %rsi, q in %rdx

aframe:

  pushq %rbp

  movq %rsp, %rbp

  subq $16, %rsp

  leaq 30(,%rdi,8), %rax

  andq $-16, %rax

  subq %rax, %rsp

  leaq 15(%rsp), %r8

  andq $-16, %r8

Explanation:

• The instruction “pushq %rbp” creates a stack element and pushes stack pointer.
• The instruction “movq %rsp, %rbp” stores stack pointer into “%rbp”.
• The instruction “subq $16, %rsp” allocates space for “i”.
  • It sets “s1” in “%rsp”.
• The instruction “leaq 30(,%rdi,8), %rax” loads value at offset “$8*\%rdi+30$ ” into register “%rax”.
• The instruction “andq $-16, %rax” performs “AND” operation between given immediate value and value in register “%rax”.
• The instruction “subq %rax, %rsp” allocates space for array “p”.
  • It sets “s2” in “%rsp”.
• The instruction “leaq 15(%rsp), %r8” loads value at offset “$\%rsp+15$ ” into register “%r8”

B.

Explanation of Solution

Logic in computation of p:

• The expression for “p” is shown below:

  $p=(s2+15)\&0$

C.

Explanation of Solution

Values of n and s₁:

• The details of values are shown below:

which	e1	n	s1
Minimum	1	even	n%16==1
Maximum	24	odd	n%16==0

• The details of minimum case are shown below:
  • The “e₁” cannot be 0, if “e₁” equals 0, “p” should be equal to “s₂”.
  • When “n” is even, “$e_1+e_2==16$”, if “e₂” is 15, “e₁” will be minimum value.
  • Hence, “$s_1==n$” and “$n\%16==1$”

D.

Explanation of Solution

Alignment properties:

• The value “p” is been aligned by 16.
• The value “s₂” denotes least multiple of 16...