Test Exam Solutions

Problem 1.
a)
  0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15 
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---
| c   X   X   X   p   p   p   p   i   i   i   i   X   X   X   X 
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---

 16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
  d   d   d   d   d   d   d   d   s   s   X   X   X   X   X   X 
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+

b) 32  c) 8  d) 5  e) 8  f) 8


Problem 2.
1) f  2) b  3) a  4) c  5) e  6) h

In all the reasoning we examplify with 4-bit numbers
a) MIN_INT + MAX_INT = 1000 + 0111 = 1111 = -1
   Now we only have to look at one bit:
   b ^ 1 = ~b so we get ~(~a | ~b) = ~~a & ~~b = a & b
   Calculates a & b
b) Look at one bit. Two cases b = 1 and b = 0
   b = 1) ((a ^ 1)  & ~1) | (~(a ^ 1) & 1) = (~a & 0) | (~~a) = a
   b = 0) ((a ^ 0)  & ~0) | (~(a ^ 0) & 0) = (a & 1) | (~a & 0) = a
   Calculates a
c) 1 + (a << 3) + ~a = (a << 3) + ~a + 1 = a * 8 - a = a * 7
   Calculates a * 7
d) (a << 4) + (a << 2) + (a << 1) = a * 16 + a * 4 + a * 2 = a * 22
   Calculates a * 22
e) Two cases a < 0 and a >= 0
   a < 0) Need to add a bias 2^2 - 1 = 3 to get closest integer >= a / 4
         (a + 3) >> 2 = a / 4
   a >= 0) a >> 2 = a / 4
   Calculates a / 4
f) See a) We get a ^ 1 = ~a
   Calculates ~a
g) ~a + 1 = -a
   Two cases a = 0 and a != 0
   a = 0) ~((0 | -0) >> W) & 1 = ~(0 >> W) & 1 = ~0 & 1 = 1
   a != 0) a is of form b 1 z where b are m arbitrary bits and z are n zeros
   where m>=0, n>=0 and m+n=W. Then ~a is ~b 0 k where k are n ones and we get
   ~b 0 k + 1 = ~b 1 z which gives b 0 z | ~b 1 z = d z where d is
   m+1 ones so ~(d z) = p k where p are m+1 zeros
   (p k) >> W = 0 and 0 & 1 = 0
   So a = 0 gets 1 and a != 0 gets 0. Calculates !a
h) Two cases a < 0 and a >= 0
   a < 0) a >> W = -1 and -1 << 1 = -2 but ~x = -x - 1 so ~(-2) = -(-2) - 1 = 2 - 1 = 1
   a >= 0) a >> W = 0 and 0 << 1 = 0 and ~0 = -1
   Calculates (a < 0) ? 1 : -1
i) Calculates a >> 2

Problem 3.
Description          |     Binary   |    M    |   E    |   Value
------------------------------------------------------------------
Minus Zero           |  1 000 0000  |    0    |   -2   |   -0.0
------------------------------------------------------------------
---                  |  0 100 0101  |  21/16  |    1   |   21/8
------------------------------------------------------------------
Smallest Denormalized|  1 000 1111  |  15/16  |   -2   |  -15/64
------------------------------------------------------------------
Largest normalized   |  0 110 1111  |  31/16  |    3   |   31/2
------------------------------------------------------------------
One                  |  0 011 0000  |    1    |    0   |   1.0
------------------------------------------------------------------
---                  |  0 101 0110  |  11/8   |    2   |   5.5 
------------------------------------------------------------------
Positive infinity    |  0 111 0000  |   ---   |   ---  | +\infty
------------------------------------------------------------------


Problem 4.
M=15   N=9


Problem 5.
fun1


Problem 6.
A) yes, -4, Need to pass pointer to it to recursive call
B) no
C) the value of %ebx is saved here, because %ebx is a callee-save register.
D) nothing is stored here.


Problem 7.
A) 1 as we remove the jle instruction.
B) if %eax <= 0 we gain 0 as we remove the jle instruction but we
   execute the rrmovlg which does nothing
   if %eax > 0 we gain 3 as we remove the jle instruction and avoid 2 bubbles
C)