INEL4206PE1sol.pdf

Problem 1: Programming in C [26 Points]
Write a program in C that does the following:
a) receives three integer numbers from the console
b) if they are already sorted in non-ascending order, it prints “Great! In n tries.” And
ends
c) if they are not already sorted in non-ascending order, it prints “Try again…”, and
goes to step a.
#include
#include
int main(int argc, char *argv[])
{
//Variable declaration
int flag; //loop condition variable
int maxIter; //max number of iterations
int first, second, third; //input integers
//variable initialization
flag=0;
maxIter=0;
while(flag!=1){
maxIter++; //incrementing the loop counter
printf("Please enter three integers in non-ascending order \n");
printf("Enter the first integer \n");
scanf("%d",&first);
printf("Enter the second integer \n");
scanf("%d",&second);
printf("Enter the third integer \n");
scanf("%d",&third);
//cheking for non-ascending order
if( (first >= second) && (second >= third) ){
printf(" %d %d %d \n",first , second ,third);
printf("Great in %d tries", maxIter);
flag=1;
}
else{
//input is not in decreasing order
printf(" %d %d %d \n",first , second ,third);
printf("Try Again......");
}
}// end while
system("PAUSE");
return 0;
}
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Problem 2: Number Systems [30 points]
a) Convert -20.125 from decimal number into single-precision floating point number:
Write your final answer here:
1 10000011 01000010000000000000000
Show the details of calculations here:
20 10100
0.125 0.001
20.125 10100.001
10100.001=1.0100001 *2^4
4+127=131
131 10000011
b) Convert +65 from decimal numbers into signed binary words.
Write your final answer here:
00000000 01000001
c) Convert -65 from decimal numbers into signed binary words.
Write your final answer here:
11111111 10111111
Show the details of calculations here:
-65= -128+32+16+8+4+2+1
11111111 10111111
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Problem 2: Number Systems (continued)
d) Convert AB.CD from hexadecimal number into binary-coded hexadecimal code (BCH)
Your answer here:
1010 1011 . 1010 1101
e) Show how 15-10 (15 minus 10) is calculated using the 2’s complement and the addition operation.
Note. You need to convert both numbers to binary first.
Your answer here:
15 – 10 = 15 + 2’s comp of (10)
15 1111
10 1010
1111 – 1010 = 1111 + 0110= 1/ 0101
Since carry bit is 1, we ignore it and the final answer is 0101
f) Convert 11011.11from binary to decimal.
Your answer here:
16+8+2+1+1/2+1/4=27.75
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Problem 3: Short Answer Questions: [24 Points]
a) What is the purpose of IP/EIP register?
b) What memory location is addressed by in the real mode 80286 register when SS=2900H and
SP=2A00H ?
2BA00H
c) Protected mode memory addressing allows access to which area of the memory in the 80286
microprocessor?
ALL 16 MB
e) How much memory is found in the Windows system area?
f) How much memory is found in the DOS transient program area?
Keeps the address of the next step of the
program
d) If DS=0107H in a protected mode system, which entry, which table, and which requested privilege level
are selected?
Entry:
Table:
Requested privilege
20H
Local
level: 3
2GB
640KB
The content of a segment register during protected mode operation of the 80286-Core2 µp
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