Computer Systems: A Programmer's Perspective (3rd Edition)
3rd Edition
ISBN: 9780134092669
Author: Bryant, Randal E. Bryant, David R. O'Hallaron, David R., Randal E.; O'Hallaron, Bryant/O'hallaron
Publisher: PEARSON
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Expert Solution & Answer
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 “
- 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 “
B.
Explanation of Solution
Logic in computation of p:
- The expression for “p” is shown below:
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, “
- Hence, “
D.
Explanation of Solution
Alignment properties:
- The value “p” is been aligned by 16.
- The value “s2” denotes least multiple of 16...
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Chapter 3 Solutions
Computer Systems: A Programmer's Perspective (3rd Edition)
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