COMPUTER SCIENCE ILLUMINATED
7th Edition
ISBN: 9781284208047
Author: Dale
Publisher: JONES+BART
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Chapter 6, Problem 34E
Explanation of Solution
Consider the given instructions:
0001 D0 00 07
0004 70 00 02
Executing the first instruction:
Consider the first instruction “D0 00 07” stored in the address “0001”.
- Here, the hexadecimal value “D0” is an instruction and “00 07” is the operand.
- The binary equivalent of the instruction “D0” is “1101 0000”, which will be stored in the instruction specifier.
- In the instruction specifier, the first four bits refers the operation code, the fifth bit refers the register specifier, and the next three bits refers the addressing mode.
- The first four bits “1101” – Load byte into the “A” register.
- The fifth bit “0” – Indicates the “A” register (Accumulator).
- The next three bits “000” – Immediate addressing mode.
- The binary equivalent of the operand “00 07” is “0000 0000 0000 0111”, which will be stored in the operand specifier.
- Since the addressing mode is immediate, the first byte of the operand specifier is ignored and the second byte value will be stored into the accumulator.
- So, the second byte value “0000 0111” will be stored into the accumulator.
Thus, after the execution of the first instruction, the “A” register will contain the value “0000 0000 0000 0111”...
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// implement a loop of your choice here
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root=number/2;
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root old root;
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} 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;
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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. 25ECh. 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. 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. 2TQCh. 6 - Prob. 3TQCh. 6 - Prob. 4TQ
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