Programming Language Pragmatics, Fourth Edition
4th Edition
ISBN: 9780124104099
Author: Michael L. Scott
Publisher: Elsevier Science
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Expert Solution & Answer
Chapter 5, Problem 3E
Explanation of Solution
- a. Converting the decimal number to hexadecimal:
- The base value of hexadecimal is 16.
- Hexadecimal value consists of 1-9 numbers and A-F letters which defines the value of 10-16.
- For the given number 1234, following steps explains the conversion of decimal to hexadecimal value...
- The base value of hexadecimal is 16.
- Hexadecimal value consists of 1-9 numbers and A-F letters which defines the value of 10-16.
- For the given number 1234, following steps explains the conversion of decimal to hexadecimal value...
Explanation of Solution
- b. Converting unsigned hexadecimal number to decimal:
- The decimal number has a base value of 10.
- In the given number the hexadecimal number a represents 10 and e represents 14.
- That is (2ae)16 =(21012)16.
- For decimal conversion, each digit of the number is to be multiplied by 16...
- The decimal number has a base value of 10.
- In the given number the hexadecimal number a represents 10 and e represents 14.
- That is (2ae)16 =(21012)16.
- For decimal conversion, each digit of the number is to be multiplied by 16...
Explanation of Solution
- c. Interpreting the hexadecimal number to 2’s complement number for the number 0xffd9:
- In the given number f represents 15 and d represents 13. So the resultant is as follows.
- In the given number f represents 15 and d represents 13. So the resultant is as follows.
(ffd9)16=(1515139)16.
- Converting the hexadecimal number to binary.
(ffd9)16 =(1111 1111 1101 1001)2...
Explanation of Solution
- d. Justification:
- • To get a negative of an integer, one can invert the bits of the number and then add 1.
- • One can do the same by subtracting the number from 0 or 2k, k is the number of bits...
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Chapter 5 Solutions
Programming Language Pragmatics, Fourth Edition
Ch. 5.2 - Prob. 1CYUCh. 5.2 - Prob. 2CYUCh. 5.2 - Prob. 3CYUCh. 5.2 - Prob. 4CYUCh. 5.2 - Prob. 5CYUCh. 5.2 - Prob. 6CYUCh. 5.3 - Prob. 7CYUCh. 5.3 - Prob. 8CYUCh. 5.3 - Prob. 9CYUCh. 5.3 - Prob. 10CYU
Ch. 5.3 - Prob. 11CYUCh. 5.3 - Prob. 12CYUCh. 5.3 - Prob. 13CYUCh. 5.3 - Prob. 14CYUCh. 5.3 - Prob. 15CYUCh. 5.3 - Prob. 16CYUCh. 5.3 - Prob. 17CYUCh. 5.3 - Prob. 18CYUCh. 5.4 - Prob. 19CYUCh. 5.4 - Prob. 20CYUCh. 5.4 - Prob. 21CYUCh. 5.4 - Prob. 22CYUCh. 5.4 - Prob. 25CYUCh. 5.4 - Prob. 27CYUCh. 5.4 - Prob. 28CYUCh. 5.4 - Prob. 29CYUCh. 5.4 - Prob. 30CYUCh. 5.4 - Prob. 31CYUCh. 5.4 - Prob. 32CYUCh. 5.4 - Prob. 33CYUCh. 5.4 - Prob. 34CYUCh. 5.4 - Prob. 35CYUCh. 5.4 - Prob. 36CYUCh. 5.4 - Prob. 37CYUCh. 5.4 - Prob. 38CYUCh. 5.5 - Prob. 39CYUCh. 5.5 - Prob. 40CYUCh. 5.5 - Prob. 41CYUCh. 5.5 - Prob. 42CYUCh. 5.5 - Prob. 43CYUCh. 5.5 - Prob. 44CYUCh. 5.5 - Prob. 45CYUCh. 5.5 - Prob. 46CYUCh. 5.5 - Prob. 47CYUCh. 5.5 - Prob. 48CYUCh. 5 - Prob. 1ECh. 5 - Prob. 2ECh. 5 - Prob. 3ECh. 5 - Prob. 4ECh. 5 - Prob. 5ECh. 5 - Prob. 6ECh. 5 - Prob. 7ECh. 5 - Prob. 8ECh. 5 - Prob. 9ECh. 5 - Prob. 10ECh. 5 - Prob. 11ECh. 5 - Prob. 13ECh. 5 - Prob. 14ECh. 5 - Prob. 15ECh. 5 - Prob. 16ECh. 5 - Prob. 17ECh. 5 - Prob. 18ECh. 5 - Prob. 19E
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