Midterm 1
Name:
Tuesday September 25, 2018
UIN:
Page 1 of 6
Midterm 1 – CS 261
Please read this page and follow the directions before
proceeding with the rest of the exam.
1) Put your name on every page of this exam in the space provided on the
top of each page. Be sure to put your name on any scratch paper you submit
with the exam. Doing so will get you 2 points.
2) Carefully read the entire exam while doing 1) above
3) Place your answers in the boxes provided to assist in grading
4) GOOD LUCK!!!
Problem
Page 2
Page 3
Page 4
Page 5
Directions
Total
Max Points
25
30
21
24
2/0
100
Points
Midterm 1
Name:
Tuesday September 25, 2018
UIN:
Page 2 of 6
1. Write a sequence of C bitwise operations that will create the 3-nibble value RESULT,
that is composed of the 0th, 3rd, and 6th nibble from the 32-bit quantity stored in the
variable X. For example, if X is equal to 0xABCDEF98 (0x8 is the 0th nibble, E is the 3rd
nibble, and B is the 6th nibble) your sequence should return 0x0BE8 in RESULT. You can
assume X is declared as an int and RESULT is declared as an int on systems3/systems4
(15 points)
2. Encode the decimal value 2018 as a signed 2's complement binary number using 16
bits. (5 points)
3. Encode the decimal value -2018 as a signed 2's complement binary number using 16
bits. (5 points)
Midterm 1
Name:
Tuesday September 25, 2018
UIN:
Page 3 of 6
4. Encode the binary value 1011110110101011100 as a hexadecimal value (5 points)
5. Give the binary for the following hexadecimal value 0x3C7A1 (5 points)
6. Assume that rbp, rax, and rcx are initialized to the values shown in the table BEFORE
each instruction is executed. State the 64-bit value that will be in rcx AFTER each
instruction is executed. (5 points each instruction) (20 points total)
rbp
rax
rcx
0x0000000000FE840 0x0000000000000003 0xFFFF0000FFFF0000
Instruction
leaq
64-bit result in rcx after the instruction is
executed
8(%rbp,%rax,8), %rcx
andq 8(%rbp), %rcx
movq
0x10(%rbp),%rcx
sarq $6, %rcx
Address
Memory
Value
0xFE830
0xDEED0DAD0FED0ADE
0xFE838
0xCAB0DAB0BED0ACE0
0xFE840
0xABABABABABABABAB
0xFE848
0x1C532145541235C1
0xFE850
0x1234567890ABCDEF
0xFE858
0xABCDEF0134561327
0xFE860
0x72AE983C4E2E3CB1
Midterm 1
Name:
Tuesday September 25, 2018
UIN:
Page 4 of 6
7. Using the values shown for the memory (some of which is the stack), $rsp, $rax and
$rbx, state which register changes as a result of each instruction, and what the new value
for the register will be after the instruction. Start with the same initial stack and register
values for each instruction. (21 points)
rsp
rax
rbx
0x0000000000FE848 0x0000000000000003 0xFFFF0000FFFF0000
Memory / Stack
Address
Value
0xFE830
0xDEED0DAD0FED0ADE
0xFE838
0xCAB0DAB0BED0ACE0
0xFE840
0xABABABABABABABAB
0xFE848
0x1C532145541235C1
0xFE850
0x1234567890ABCDEF
0xFE858
0xABCDEF0134561327
0xFE860
0x72AE983C4E2E3CB1
Which register
changes
(2 pts each)
a) pushq $rax
b) movq 8($rsp), $rbx
c) ret
New register value is
(5 pts each)
Midterm 1
Name:
Tuesday September 25, 2018
UIN:
Page 5 of 6
7. For each assembly code snippet, give the C code that could have produced the
assembly when compiled with gcc. For variables, i is stored at -4(%rbp) and j is stored at
-8(%rbp). (24 points total, 8 points each snippet)
Assembly
C code that produces the assembly to
the left
movl
jmp
$5, -4(%rbp)
.L2
.L3:
** some assembly code **
addl
$2, -4(%rbp)
.L2:
cmpl
$14, -4(%rbp)
jle
.L3
movl
movl
cmpl
jg
-4(%rbp), %edx
-8(%rbp), %eax
%eax, %edx
.L2
** some assembly code **
jmp
.L3
.L2:
** some assembly code **
.L3:
sarl
movl
addl
movl
addl
movl
-8(%rbp)
-8(%rbp), %eax
%eax, -4(%rbp)
-4(%rbp), %eax
%eax, %eax
%eax, -8(%rbp)
Midterm 1
Name:
Tuesday September 25, 2018
UIN:
Page 6 of 6