Solutions to End of Chapter Problems
Problem 1.
(a)
By the presence of the NULL character (0x00) which signifies the end of the string.
(b)
This string can hold 19 characters (allowing a space for the NULL). There are already 14
characters in the string… so there is room for 5 more.
Problem 2.
(a) S
(b) S
(c) U
(d) U
Problem 3.
(a)
0x080483AC
(b)
The hexadecimal number stored is 0x7266DB05. This equates to the decimal number
1,919,343,365
(c)
0x080483A0
(d)
0x080483A8
(e)
BRAVO!
Problem 4.
(e)
Problem 5.
My teacher is LCDR Agood day by all!
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