Instructor’s Solution Manual
Computer Systems:
A Programmer’s Perspective
1
,2
Chapter 1
Solutions to Homework Problems
The text uses two different kinds of exercises:
Practice Problems. These are problems that are incorporated directly into the text, with explanatory
solutions at the end of each chapter. Our intention is that students will work on these problems as they
read the book. Each one highlights some particular concept.
Homework Problems. These are found at the end of each chapter. They vary in complexity from
simple drills to multi-week labs and are designed for instructors to give as assignments or to use as
recitation examples.
This document gives the solutions to the homework problems.
1.1 Chapter 1: A Tour of Computer Systems
1.2 Chapter 2: Representing and Manipulating Information
Problem 2.40 Solution:
This exercise should be a straightforward variation on the existing code.
code/data/show-ans.c
1 void show_short(short int x)
2 {
3 show_bytes((byte_pointer) &x, sizeof(short int));
4 }
5
6 void show_long(long int x)
7 {
8 show_bytes((byte_pointer) &x, sizeof(long));
9 }
1
,
, 2 CHAPTER 1. SOLUTIONS TO HOMEWORK PROBLEMS
10
11 void show_double(double x)
12 {
13 show_bytes((byte_pointer) &x, sizeof(double));
14 }
code/data/show-ans.c
Problem 2.41 Solution:
There are many ways to solve this problem. The basic idea is to create some multibyte datum with different
values for the most and least-significant bytes. We then read byte 0 and determine which byte it is.
In the following solution is to create an int with value 1. We then access its first byte and convert it to an
int. This byte will equal 0 on a big-endian machine and 1 on a little-endian machine.
code/data/show-ans.c
1 int is_little_endian(void)
2 {
3 /* MSB = 0, LSB = 1 */
4 int x = 1;
5
6 /* Return MSB when big-endian, LSB when little-endian */
7 return (int) (* (char *) &x);
8 }
code/data/show-ans.c
Problem 2.42 Solution:
This is a simple exercise in masking and bit manipulation. It is important to mention that ˜0xFF is a way
to generate a mask that selects all but the least significant byte that works for any word size.
(x & 0xFF) | (y & ˜0xFF)
Problem 2.43 Solution:
These exercises require thinking about the logical operation ! in a nontraditional way. Normally we think
of it as logical negation. More generally, it detects whether there is any nonzero bit in a word.
A. !!x
B. !!˜x
C. !!(x & 0xFF)
D. !!(˜x & 0xFF)
Problem 2.44 Solution:
Computer Systems:
A Programmer’s Perspective
1
,2
Chapter 1
Solutions to Homework Problems
The text uses two different kinds of exercises:
Practice Problems. These are problems that are incorporated directly into the text, with explanatory
solutions at the end of each chapter. Our intention is that students will work on these problems as they
read the book. Each one highlights some particular concept.
Homework Problems. These are found at the end of each chapter. They vary in complexity from
simple drills to multi-week labs and are designed for instructors to give as assignments or to use as
recitation examples.
This document gives the solutions to the homework problems.
1.1 Chapter 1: A Tour of Computer Systems
1.2 Chapter 2: Representing and Manipulating Information
Problem 2.40 Solution:
This exercise should be a straightforward variation on the existing code.
code/data/show-ans.c
1 void show_short(short int x)
2 {
3 show_bytes((byte_pointer) &x, sizeof(short int));
4 }
5
6 void show_long(long int x)
7 {
8 show_bytes((byte_pointer) &x, sizeof(long));
9 }
1
,
, 2 CHAPTER 1. SOLUTIONS TO HOMEWORK PROBLEMS
10
11 void show_double(double x)
12 {
13 show_bytes((byte_pointer) &x, sizeof(double));
14 }
code/data/show-ans.c
Problem 2.41 Solution:
There are many ways to solve this problem. The basic idea is to create some multibyte datum with different
values for the most and least-significant bytes. We then read byte 0 and determine which byte it is.
In the following solution is to create an int with value 1. We then access its first byte and convert it to an
int. This byte will equal 0 on a big-endian machine and 1 on a little-endian machine.
code/data/show-ans.c
1 int is_little_endian(void)
2 {
3 /* MSB = 0, LSB = 1 */
4 int x = 1;
5
6 /* Return MSB when big-endian, LSB when little-endian */
7 return (int) (* (char *) &x);
8 }
code/data/show-ans.c
Problem 2.42 Solution:
This is a simple exercise in masking and bit manipulation. It is important to mention that ˜0xFF is a way
to generate a mask that selects all but the least significant byte that works for any word size.
(x & 0xFF) | (y & ˜0xFF)
Problem 2.43 Solution:
These exercises require thinking about the logical operation ! in a nontraditional way. Normally we think
of it as logical negation. More generally, it detects whether there is any nonzero bit in a word.
A. !!x
B. !!˜x
C. !!(x & 0xFF)
D. !!(˜x & 0xFF)
Problem 2.44 Solution:
