Computer Science Illuminated
6th Edition
ISBN: 9781284055917
Author: Nell Dale, John Lewis
Publisher: Jones & Bartlett Learning
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Chapter 6, Problem 34E
Explanation of Solution
Execution of the first instruction “0001 31 00 0F”:
Given instruction, 0001 31 00 0F
- The value “0001” is the address of the instruction.
- The hexadecimal value “31” is an instruction and “00 0F” is the operand.
- The mnemonic for the hexadecimal value “31” is “DECI”.
- The binary value for the hexadecimal instruction “31” is “0011 0001”.
- The first five bits express the meaning of the instruction.
- “00110” – Decimal input to the operand “(DECI)”.
- So, the instruction “31” gets the decimal input from the user.
- Enter “5” as input.
- The last three bits are used as addressing mode specifier.
- “001” – Direct addressing mode.
- The direct addressing mode represents the address of data in the operand.
- “001” – Direct addressing mode.
- So, it loads the input value “5” to the given operand’s location “00 0F”.
Address and its values
| Address | Value |
| 000F | 5 |
Execution of the second instruction “0004 C1 00 0F”:
Given instruction, 0004 C1 00 0F
- The value “0004” is the address of the instruction.
- The hexadecimal value “C1” is an instruction, and “00 0F” is the operand.
- The mnemonic for the hexadecimal value “C1” is “LDr”.
- The binary value for the hexadecimal instruction “C1” is “1100 0001”.
- The first four bits express the meaning of the instruction.
- “1100” – Load the operand into the A register “(LDr)”.
- The last three bits are used as addressing mode specifier.
- “001” – Direct addressing mode.
- The direct addressing mode represents the address of data in the operand.
- Here, the value in the address represented in the operand “00 0F” is “5”.
- “001” – Direct addressing mode.
- So, it loads the value “5”, from the address stored in the operand “00 0F”, into the accumulator...
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public class SquareRoot {
public static void main(String[] args) {
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// implement a loop of your choice here
// Method that calculates the square root of integer variables
public static double myRoot(int number) {
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root=number/2;
double root old;
do {
root old root;
root (root_old+number/root_old)/2;
} while (Math.abs(root_old-root)>1.8E-6);
return root;
}
// Method that calculates the square root of double variables
public static double myRoot(double number) {
double root;
root number/2;
double root_old;
do {
root old root;
root (root_old+number/root_old)/2;
while (Math.abs (root_old-root)>1.0E-6);
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Chapter 6 Solutions
Computer Science Illuminated
Ch. 6 - Prob. 1ECh. 6 - Prob. 2ECh. 6 - Prob. 3ECh. 6 - Prob. 4ECh. 6 - Prob. 5ECh. 6 - Prob. 6ECh. 6 - Prob. 7ECh. 6 - Prob. 8ECh. 6 - Prob. 9ECh. 6 - Prob. 10E
Ch. 6 - Prob. 11ECh. 6 - Prob. 12ECh. 6 - Prob. 13ECh. 6 - Prob. 14ECh. 6 - Prob. 15ECh. 6 - Prob. 16ECh. 6 - Prob. 17ECh. 6 - Prob. 18ECh. 6 - Prob. 19ECh. 6 - Prob. 20ECh. 6 - Prob. 21ECh. 6 - Prob. 22ECh. 6 - Prob. 23ECh. 6 - Prob. 24ECh. 6 - Prob. 26ECh. 6 - Prob. 27ECh. 6 - Prob. 28ECh. 6 - Prob. 29ECh. 6 - Prob. 30ECh. 6 - Prob. 31ECh. 6 - Prob. 32ECh. 6 - Prob. 33ECh. 6 - Prob. 34ECh. 6 - Prob. 35ECh. 6 - Prob. 36ECh. 6 - Prob. 37ECh. 6 - Prob. 38ECh. 6 - Prob. 39ECh. 6 - Prob. 40ECh. 6 - Prob. 41ECh. 6 - Prob. 42ECh. 6 - Prob. 43ECh. 6 - Prob. 44ECh. 6 - Prob. 45ECh. 6 - Prob. 46ECh. 6 - Prob. 47ECh. 6 - Prob. 48ECh. 6 - Prob. 49ECh. 6 - Prob. 50ECh. 6 - Prob. 51ECh. 6 - Prob. 52ECh. 6 - Prob. 53ECh. 6 - Prob. 54ECh. 6 - Prob. 55ECh. 6 - Prob. 56ECh. 6 - Prob. 57ECh. 6 - Prob. 58ECh. 6 - Prob. 59ECh. 6 - Prob. 60ECh. 6 - Prob. 1TQCh. 6 - Prob. 3TQCh. 6 - Prob. 4TQ
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