Againnoticethatthe2leftmostcarrybitsatthetoparediffere

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0 0 0 0 1 0 0 1 0 (‐46)10 (010010)2s = (+20)10 ≠ (‐46)10 Overflow must have occurred. Again notice that the 2 leftmost carry bits at the top are different. c) 0 0 1 1 (‐25)10 (‐0 1 1 0 0 1) 1 0 0 1 1 1 (18)10 + 0 1 0 0 1 0 = + 0 1 0 0 1 0 1 1 1 0 0 1 (‐7)10 (111001)2s = ‐(000111)2 = (‐7)10 The result is correct. Notice that the 2 leftmost carry bits at the top are the same. d) 1 1 1 1 (‐12)10 (‐0 0 1 1 0 0) 1 1 0 1 0 0 (13)10 + 0 0 1 1 0 1 = + 0 0 1 1 0 1 0 0 0 0 0 1 (1)10 (000001)2s = (1)10 The result is correct. Notice that the 2 leftmost carry bits at the top are the same. e) 1 1 1 1 1 1 (‐11)10 (‐0 1 0 1 0 1) 1 0 1 0 1 1 (‐21)10 + (‐0 0 1 0 1 1) = + 1 1 0 1 0 1 1 0 0 0 0 0 (‐32)10 (100000)2s = (‐32)10 The result is correct. Notice that the 2 leftmost carry bits at the top are the same. 0 1 1 1 1 4) a) Find the value of z for the following problems using bit‐shifting. Truncate the remainder if applicable. All values represent unsigned numbers. i) (01101001)2 / (1000)2 = (00001101)2 ii) (1010)2 x (0010)2 = (00010100)2 iii) (A43F)16 / (0004)16 = (1010 0100 0011 1111)2 / (0000 0000 0000 0100)2 = (0010 1001 0000 1111)2 32‐bit => 8‐hex digits = (0000 290F)16 iv) (FFFF)16 x (20)16 = (1111 1111 111...
View Full Document

Ask a homework question - tutors are online