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8.1 Among the 13 declarations: a. The following are valid declarations for a C++ character string: char s[6]; char s[6] = {'H', 'e', 'l', 'l', 'o'}; char s[6] = "Hello"; char s[] = {'H', 'e', 'l', 'l', 'o'}; char s[] = "Hello"; char* s;
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[CHAP. 8
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char* s = new char[6]; char* s = "Hello"; Warning: this last declaration only defines s to be a pointer to a string constant. b. The following are valid declarations for a C++ character C-string of length 5, initialized to the C-string "Hello" and allocated at compile time: char s[6] = {'H', 'e', 'l', 'l', 'o'}; char s[6] = "Hello"; char s[] = {'H', 'e', 'l', 'l', 'o'}; char s[] = "Hello"; char* s = "Hello"; // defines s as a pointer to a string constant c. It is not possible to initialize a C-string like this at run time. d. The following are valid declarations for a C++ character string as a formal parameter for a function: char s[]; char* s; This will read only as far as the first whitespace. For the given input, it would assign "Hello," to s. This counts the number of uppercase letters in the C-string s, so the output is 6. This changes all uppercase letters to lowercase in the C-string s: 123 w. 42nd st., ny, ny 10020-1095 Note that to change the case of a character *p, it must be assigned the return value of the function: *p = tolower(*p); This increments all uppercase letters, changing the W to an X, the S to a T, etc.: 123 X. 42nd Tt., OZ, OZ 10020-1095 This counts the number of punctuation characters in the C-string s, so the output is 5. It changes each character that is followed by a punctuation character to that following character: 123 .. 42nd S.,, N,, NY 1002--1095 The assignment s1 = s2 simply makes s1 a synonym for s2; i.e., they both point to the same character. The call strcpy(s1,s2) actually copies the characters of s2 into the C-string s1, thereby duplicating the C-string. a. This assigns the integer 10 to n. b. This assigns the substring "rford" to s1. c. This assigns the substring "rd" to s1. d. This assigns the substring "utherford" to s1. e. This copies last to first, so that first will also be the string "Hayes". f. This copies the substring "Hay" into the first part of first, making it "Hayherford". g. This appends last onto the end of first, making it "RutherfordHayes". h. This appends the substring "Hay" onto the end of first, making it "RutherfordHay". a. 7. b. 6. c. 5. d. 7. It prints: ABCDE >= ABC It prints: ABCDE < ABCE It prints: ABCDE >= It prints: !=
Solutions to Problems
8.1 This does not work because the assignment name[count] = buffer;
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CHAP. 8]
C-STRINGS
assigns the same pointer to each of the C-strings name[0], name[1], etc. Arrays cannot be assigned this way. To copy one array into another, use strcpy(), or strncpy(). This copies the C-string s2 into the C-string s1: char* strcpy(char* s1, const char* s2) { char* p; for (p=s1; *s2; ) *p++ = *s2++; *p = '\0'; return s1; } The pointer p is initialized at the beginning of s1. On each iteration of the for loop, the character *s2 is copied into the character *p, and then both s2 and p are incremented. The loop continues until *s2 is 0 (i.e., the null character '\0'). Then the null character is appended to the C-string s1 by assigning it to *p. (The pointer p was left pointing to the byte after the last byte copied when the loop terminated.) Note that this function does not allocate any new storage. So its first argument s1 should already have been defined to be a character string with the same length as s2. This function appends up to n characters from s2 onto the end of s1. It is the same as the strcat() function except that its third argument n limits the number of characters copied: char* strncat(char* s1, const char* s2, size_t n) { char* end; for (end=s1; *end; end++) // find end of s1 ; char* p; for (p=s2; *p && p-s2<n; ) *end++ = *p++; *end = '\0'; return s1; } The first for loop finds the end of C-string s1. That is where the characters from C-string s2 are to be appended. The second for loop copies characters from s2 to the locations that follow s1. Notice how the extra condition p-s2<n limits the number of characters copied to n: the expression p-s2 equals the number of characters copied because it is the difference between p (which points to the next character to be copied) and s2 (which points to the beginning of the C-string). Note that this function does not allocate any new storage. It requires that C-string s1 have at least k more bytes allocated, where k is the smaller of n and the length of C-string s2. This requires testing the last letter and the second from last letter of the word to be pluralized. We use pointers p and q to access these letters. void pluralize(char* s) { int len = strlen(s); char* p = s + len - 1; // last letter char* q = s + len - 2; // last 2 letters if (*p == 'h' && (*q == 'c' || *q == 's')) strcat(p, "es"); else if (*p == 's') strcat(p, "es"); else if (*p == 'y') if (isvowel(*q)) strcat(p, "s"); else strcpy(p, "ies"); else if (*p == 'z') if (isvowel(*q)) strcat(p, "zes"); else strcat(p, "es"); else strcat(p, "s"); } Two of the tests depend upon whether the second from last letter is a vowel, so we define a little boolean function isvowel() for testing that condition: bool isvowel(char c) { return (c=='a' || c=='e' || c=='i' || c=='o' || c=='u'); }
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