D the saved value of register ebp was changed to

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Unformatted text preview: inary conversion. It also gets you thinking about integer and floating-point representations. We will explore these representations in more detail later in this chapter. A. Using the notation of the example in the text, we write the two strings as 0 0 3 5 4 3 2 1 00000000001101010100001100100001 ********************* 4 A 5 5 0 C 8 4 01001010010101010000110010000100 B. With the second word shifted two positions relative to the first we find a sequence with 21 matching bits. C. We find all bits of the integer embedded in the floating point number, except for the most signficant bit having value 1. Such is the case for the example in the text as well. In addition the floating-point number has some nonzero high-order bits that do not match those of the integer. Problem 2.4 Solution: [Pg. 33] It prints 41 42 43 44 45 46. Recall also that the library routine strlen does not count the terminating null character, and so show_bytes printed only through the character ‘F.’ Problem 2.5 Solution: [Pg. 36] This problem is a drill to help you become more fami...
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