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Unformatted text preview: 9 February 2011 CSE-1520R Test #1 [ B4 ] 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 9 February 2011 CSE-1520R Test #1 [ B4 ] p. 2 of 8 1. (10 points) Binary & Number Systems a. (2 points) You see the byte 01101100 . You know it represents A. the ASCII character l (lowercase L). B. the natural number 108. C. the negative integer- 20 D. the floating point number 3 . 0. 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 bitsay, 1 means positive and 0 means negativeand 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. one would have two different representations. E. zero would have two different representations. c. (4 points) Fill in the blanks. byte format value 8-bit floating point 6 1 2 00110010 2 11111110 8-bit signed integer, 2s complement 10100110 hexadecimal d. (2 points) Consider storing signed integers in 16 bits in twos complement format. Thed....
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- Spring '11