COMPUTER SYSTEMS&MOD MSGT/ET SA AC PKG
3rd Edition
ISBN: 9780134671123
Author: Bryant
Publisher: PEARSON
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Expert Solution & Answer
Chapter 4.1, Problem 4.4PP
Explanation of Solution
Given C code:
long rsum(long *start, long count)
{
if(count <= 0)
return 0;
return *start + rsum(start+1, count-1);
}
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.
- movb:
Corresponding x86-64 code:
long rsum(long *start, long count)
start in %rdi, count in %rsi
rsum:
movl $0, %eax
testq %rsi, %rsi
jle .L9
pushq %rbx
movq (%rdi),%rbx
subq $1, %rsi
addq $8, %rdi
call rsum
addq %rbx, %rax
popq %rbx
.L9:
rep; ret
Explanation:
- The instruction “movl $0, %eax” initializes value in register “%eax” to 0.
- The instruction “testq %rsi, %rsi” checks for count to be zero.
- The instruction “jle .L9” jumps to label “.L9” if first value is less than or equal to second.
- The instruction “pushq %rbx” saves callee-saved register.
- The instruction “movq (%rdi),%rbx” moves value at location of register “%rdi” to register “%rbx”.
- The instruction “subq $1, %rsi” decrements value of “count” stored in register “%rsi”...
Expert Solution & Answer
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Check out a sample textbook solutionStudents have asked these similar questions
The programming language: C++
The union of two ordered lists (Sequential linear list)[the solution
introduction: The first video of in 3.1, 35:00-43:00]
[Problem description]
Give the union of two ordered lists. The maximal number of elements in an
inputted set is 30.
[Basic requirements]
1) Use sequential linear list.
2) The result list should also be ordered.
[Example]
Problem: Give the union of the ordered lists (3,4,9,100,103) and
(7,9,43,53,102,105).
What you need to show in the terminal(the back part is outputted by you
and the blue part is inputted by the user, i.e., teacher):
Please input the first ordered list: (3,4,9,100,103)Please input the second ordered list: (7,9,43,53,102,105)
The union is:
(3,4,7,9,9,43,53,100,102,103,105)
(Base Indexed Memory Addressing Mode) Only at [{DS | SS | ES}: {SI | DI} + {BX +
BP}]
BA
EA
There is a combination between (Base and Index) in Effective Add. (EA)
Ex.
Described each line and write down the equation of physical add. For the following assembly
code:
01 org 100
02 MOV [BX+DI], CL
03 MOV CH, [вх+SI)
04 MOV AH, [BP+DI]
05 MOV [BP+SI], AL
06 ret
[1] (
Show your work. Show hoe you compute memory address by using the effective memory
address computation.
Assume the following values are stored at the indicated memory addresses and
registers:
Address Value
0x100
OxFF
0x104
OxAB
0x108
0x13
0x10c
0x11
Register
%rax
%rcx
%rdx
$0x108
(%rax)
4(%rax)
9(%rax, %rdx)
260(%rcx,%rdx)
OxFC (,%rcx, 4)
(%rax, %rdx, 4)
Value
0x100
0x1
0x3
Fill in the following table showing the values for the indicated operands:
Operand
Value
%rax
0x104
Chapter 4 Solutions
COMPUTER SYSTEMS&MOD MSGT/ET SA AC PKG
Ch. 4.1 - Prob. 4.1PPCh. 4.1 - Prob. 4.2PPCh. 4.1 - Prob. 4.3PPCh. 4.1 - Prob. 4.4PPCh. 4.1 - Prob. 4.5PPCh. 4.1 - Prob. 4.6PPCh. 4.1 - Prob. 4.7PPCh. 4.1 - Prob. 4.8PPCh. 4.2 - Practice Problem 4.9 (solution page 484) Write an...Ch. 4.2 - Prob. 4.10PP
Ch. 4.2 - Prob. 4.11PPCh. 4.2 - Prob. 4.12PPCh. 4.3 - Prob. 4.13PPCh. 4.3 - Prob. 4.14PPCh. 4.3 - Prob. 4.15PPCh. 4.3 - Prob. 4.16PPCh. 4.3 - Prob. 4.17PPCh. 4.3 - Prob. 4.18PPCh. 4.3 - Prob. 4.19PPCh. 4.3 - Prob. 4.20PPCh. 4.3 - Prob. 4.21PPCh. 4.3 - Prob. 4.22PPCh. 4.3 - Prob. 4.23PPCh. 4.3 - Prob. 4.24PPCh. 4.3 - Prob. 4.25PPCh. 4.3 - Prob. 4.26PPCh. 4.3 - Prob. 4.27PPCh. 4.4 - Prob. 4.28PPCh. 4.4 - Prob. 4.29PPCh. 4.5 - Prob. 4.30PPCh. 4.5 - Prob. 4.31PPCh. 4.5 - Prob. 4.32PPCh. 4.5 - Prob. 4.33PPCh. 4.5 - Prob. 4.34PPCh. 4.5 - Prob. 4.35PPCh. 4.5 - Prob. 4.36PPCh. 4.5 - Prob. 4.37PPCh. 4.5 - Prob. 4.38PPCh. 4.5 - Prob. 4.39PPCh. 4.5 - Prob. 4.40PPCh. 4.5 - Prob. 4.41PPCh. 4.5 - Prob. 4.42PPCh. 4.5 - Prob. 4.43PPCh. 4.5 - Prob. 4.44PPCh. 4 - Prob. 4.45HWCh. 4 - Prob. 4.46HWCh. 4 - Prob. 4.47HWCh. 4 - Prob. 4.48HWCh. 4 - Modify the code you wrote for Problem 4.47 to...Ch. 4 - In Section 3.6.8, we saw that a common way to...Ch. 4 - Prob. 4.51HWCh. 4 - The file seq-full.hcl contains the HCL description...Ch. 4 - Prob. 4.53HWCh. 4 - The file pie=full. hcl contains a copy of the PIPE...Ch. 4 - Prob. 4.55HWCh. 4 - Prob. 4.56HWCh. 4 - Prob. 4.57HWCh. 4 - Our pipelined design is a bit unrealistic in that...Ch. 4 - Prob. 4.59HW
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