By Dr. Wilfred Funk, Norman Lewis
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0b0111 1111 1000 1111 0000 1111 0000 0000 b. 0b0000 0000 0000 0000 0000 0000 0000 0000 c. 0b0100 0010 0100 0000 0000 0000 0000 0000 d. 0b1000 0000 0100 0000 0000 0000 0000 0000 e. 0b1111 1111 1000 0000 0000 0000 0000 0000 Solution a. b. c. d. e. The exponent field of this number is all 1s, and its fraction field is not 0, so it’s a NaN. This number has an exponent field of 0, a sign bit of 0, and a fraction field of 0, which is the IEEE floating-point representation for þ0. This number has an exponent field of 132 and a fraction field of 100 0000 0000 0000 0000 0000.
12 b. 147 c. 384 d. 1497 Solution a. b. c. d. 12 is greater than 23 À 1, and less than 24 À 1, so 12 cannot be represented in a 3-bit unsigned integer, but can be represented in a 4-bit binary integer. Therefore, 4 bits are required. 27 À 1 < 147 < 28 À 1, so 8 bits are required. 28 À 1 < 384 < 29 À 1, so 9 bits are required. 210 À 1 < 1497 < 211 À 1, so 11 bits are required. 10. What are the largest and smallest integers representable in 4-, 8-, and 16-bit values using a. Unsigned binary representation b.
11. Convert the following decimal numbers to 8-bit sign-magnitude representation: a. 23 b. À23 c. À48 d. À65 39 40 CHAPTER 2 Data Representations Solution a. b. c. d. In sign-magnitude representation, positive integers are represented in the same way as they are in unsigned binary representation, except that the high bit of the representation is reserved for the sign bit. Therefore, the 8-bit sign-magnitude representation of 23 is 0b00010111. To get the sign-magnitude representation of À23, we simply set the sign bit of the representation of þ23 to 1, giving 0b10010111.