HW1-Solutions

HW1-Solutions - 3(0d = 0000 0000 0000 0000b 4-1024d =(1000...

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EC327 Introduction to Software Engineering HW #1 –Due Monday Sept. 21, 2009 Answered By Jiaxi Jin P1 (2 points each) 1. 0x55AA = 0101 0101 1010 1010b 2. 1 1010 1100 1110b = Ox1ACE 3. 1025 = 1024 + 1 = 2 10 + 2 0 = 100 0000 0001b 4. F00D = 46 30 30 44 (ASCII) 5. 0x3A 2D 29 = :-) P2 (2 points each) 1. (100d)* = 100d = 0000 0000 0110 0100b 2. ( 255d)* = 2 16 d – 255d (Definition of 2’s complement, 16 is the number of bits) = 65281d = 1111 1111 0000 0001b OR (-255d)* = (1000 0000 1111 1111b)* (decimal-binary transformation, first “1” is sign digit) = 1111 1111 0000 0000b + 1
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= 1111 1111 0000 0001b
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Unformatted text preview: 3. (0d)* = 0000 0000 0000 0000b 4. (-1024d)* = (1000 0100 0000 0000b)* = 1111 1011 1111 1111b + 1 = 1111 1100 0000 0000b 5. (1025d)* = 1025d = 0000 0100 0000 0001b P3 (2 points each) 1. 10 0111b + 10b = 10 1001b = 0x0000 0029 (1 point) No Overflow/OF=0 (1 point) 2. 0000 1010b + 0x10 = 0x0A + 0x10 = 0x0000 001A (1 point) No Overflow/OF=0 (1 point) 3. 0xFEED + 0xF00D = 0x0001 EEFA (1 point) OF=1 (1 point) 4. 297d + 65,239d = 65536d = 1 0000 0000 0000 0000b = 0x0001 0000 (1 point) OF=1 (1 point) 5. 0xAAAA + 0x5556 = 0x0001 0000 (1 point) OF=1 (1 point)...
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