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Solutions to End of Chapter Problems
Problem 1.
Problem 2.
The program prints Go Navy! four times, with each exciting motivational phrase on its own
line. See below:
Problem 3.
The program prints What's for lunch? and then, for reasons that defy logic, answers that
Navy is served for lunch (a double helping in fact). See below:
Problem 4.
(a)
Assembly language
(b)
This instruction takes the 4 byte value 0x08048484 and stores it in the memory
location that is stored in the esp register.
Phrased another way: The CPU first examines the address stored in the esp register.
Let's say this address stored in the esp register is the value a (for example, perhaps a is
0xbffff800). The CPU then stores the value 0x08048484 in main memory
starting at address a .
Problem 5.
(a)
(b)
(c)
The address of the next instruction to be executed
0x9b
"Yes"
6
Problem 6.
In this case, the loop iterates four times. So, the program prints I love cyber! four times,
with each statement of affection on a new line:
I
I
I
I
Problem 7.
love
love
love
love
cyber!
cyber!
cyber!
cyber!
This causes an infinite loop since the Boolean expression will always be true. The program will
print I love cyber! forever.
To terminate an infinite loop, enter Control-c
Problem 8.
#include<stdio.h>
int main ()
{
int value;
printf( "Enter an integer: ");
scanf("%d" , &value );
if(
{
value < 0
)
value = value * -1 ;
}
printf("The absolute value of the number is: %d\n" , value );
}
Problem 9.
(a)
i r eip shows that the address 0x8048384 is stored in the instruction pointer.
(b)
Looking at the assemble code, the instruction at address 0x8048384 is
mov
DWORD PTR [ebp-4],0x0
(c)
i r ebp shows that the address 0xbffff818 is stored in the base pointer.
(d)
0xbffff814
(e)
If we enter x/xw 0xbffff814 we see that the value stored is 0x00000000 .
(f)
It has changed to 0x804838b. The instruction pointer is automatically updated to hold
the address of the next instruction to be executed.
(g)
The CPU is comparing the value of i to the value of 9. The next instruction is a
"jump if less than or equal" instruction. These two assembly language instructions
collectively check to see if i  9 . Since i is an integer, checking if i  9 is the same as
checking if i < 10.
7
Problem 10.
(h)
i r eip shows that the address 0x8048393 is stored in the instruction pointer. So
we did jump.
(i)
Since 0 is less than or equal to 9, we satisfied the condition to jump to instruction
0x8048393.
(j)
0xbffff810
(k)
x/xw 0xbffff810 shows that 0x08048484 is stored at this location (as
expected!).
(l)
Entering x/s 0x08048484 we see that the string stored at this location is "Hello
World!\n".
(a)
The loop will iterate twice.
(b)
The output is:
Howitzer
Torpedoes
Problem 11.
(a)
mov DWORD PTR [ebp-4],0x5
(b)
ebp-4
(c)
0x0804838b
=
0xbffff814
8