Download Mid-term 1 sample problem solutions

CS 351 Computer Architecture
Fall 2005
Practice Problems for Mid-Semester # 1 with solutions:
1) (a) Translate the following assembly code back to equivalent C code:
Loop:
addi
addi
sle
beq
mul
addi
j
$t0,
$v0,
$t1,
$t1,
$v0,
$t0,
Loop
$zero, 1
$zero, 1
$t0, $a0
$zero, Exit
$v0, $t0
$t0, 1
#
#
#
#
#
#
#
i = 1
v = 1
set $t1 to 1 if (i < arg)
exit loop if (i >= arg)
v *= i
i ++
loop
Solution :
int v = 1;
for (int i = 1 ; i < arg ; i ++) {
v = v * i;
}
(b) What function does the above code perform?
Solution : factorial
2) Add a comment to each statement of the following MIPS assembly
procedure :
Describe the computation performed by this code.
Solution:
3) (a) Write a recursive procedure binomial(n, m) (in C) where n and m are 32 bit
n
positive integers to compute the binomial coefficient   using the formula
m
 n   n  1  n  1 
   
  
 . Make sure that you provide appropriate exit conditions.
 m   m   m  1
Solution:
int binomial(int n, int m) {
if ((m == 0) || (n == m)) return 1;
if (n == 0) return 0;
else return binomial(n-1, m) + binomial(n-1, m-1);
}
(b) Translate your code C for binomial(n, m) into MIPS assembly code.
Presented in class.
4) Shown below are five single precision floating-point integers in the IEEE 754
format. Arrange them from the smallest to the largest.
Solution: C, A, E, B, D where the numbers are A, B, C, D, E
from top to bottom.
5) Design a circuit of depth O(log n) that takes n input bits, and outputs 1 (0) if there
is an odd (even) number of 1’s in the input.



The above is the circuit for n = 4, and generalization is
obvious.
6) Design a circuit of depth O(log n) that takes n input bits b1 b2… bn and outputs the
binary representation of b1 b2… bn +1. (When all the input bits are 1, the output
should be a string of n zeroes. (i.e., overflow is ignored.) Use a recursive approach
to design the circuit and show clearly that the depth of the circuit is O(log n).
Solution: To make this recursive, we will generalize this
problem slightly. We assume that there is an extra input bit e
(enable) such that when e = 1, the output is input + 1, and
when e = 0, the output = the input. (Also, we will assume that
when the input is a string of 1’s, the output will be a string
of 0’s.
The recursive construction was presented in class.
7) Exercise 3.11
Solution:
8) Exercise 3.14
Solution:
9) Write a code segment in MIPS assembly language that sets $t0 to 1 (0) if the
architecture uses big-endian (little-endian) notation.
Solution:
li
addi
sw
lb
addi
$t0,
$sp,
$t0,
$t0,
$sp,
0x01000000
$sp, -4
0($sp)
$sp
$sp, 4
10) Exercise 3.13
Solution:
11) Exercise 2.31
Assume that the code from Ex 2.30 is run on a machine with 2 GHz clock that requires
the following number of cycles for each instruction:
add, addi, sll -> 1 cycle, lw, bne -> 2 cycles. In the worst-case, how many seconds will
it take to execute the code?
Solution:
12) Exercise 2.43
Solution:
