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# test1-7F - 9 February 2011 CSE-1520R Test#1[7F p 1 of 8...

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9 February 2011 CSE-1520R Test #1 [ 7F ] p. 1 of 8 CSE-1520R Test #1 Sur / Last Name: Given / First Name: Student ID: Instructor: Parke Godfrey Exam Duration: 45 minutes Term: Winter 2011 The exam is closed book, closed notes, and no aids such as calculators, cellphones, etc. There are five parts, each with questions. Points for each question are as indicated. Each question is multiple choice, true/false, or fill in the blank, as indicated. For multiple choice, choose the one best answer. There is no negative penalty for a wrong answer. Assume that any number you see is in decimal (base 10), unless it is clear otherwise. The test is out of 50 points. Marking Box 1. /10 2. /10 3. /10 4. /10 5. /10 Total /50

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9 February 2011 CSE-1520R Test #1 [ 7F ] p. 2 of 8 1. (10 points) Binary & Number Systems a. (2 points) You see the byte 01101010 . You know it represents A. the ASCII character ‘j’. B. the natural number 106. C. the negative integer - 22 D. the floating point number 2 . 5. E. There is not enough information to determine. b. (2 points) A natural way to “store” positive and negative (signed) integers in a byte would seem be to use the most significant bit as a sign bit—say, 1 means positive and 0 means negative—and encode the magnitude of the integer in the remaining seven bits in direct binary representation. This is not done, though, because A. we are not allowed to use the most significant bit ever. B. this encodes twice as many positive integers as negative, which is awkward. C. addition would be impossible with this format. D. zero would have two different representations. E. one would have two different representations. c. (2 points) Consider storing signed integers in 16 bits in two’s complement format. The largest integer that can be represented is A. 127 B. 128 C. 32,767 D. 32,768 E. 65,535 F. 65,536 d. (4 points) Fill in the blanks.
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