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7.15 Explain why the following alternative to Example 7.12 does not work:
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main0 char name[lO] [20], buffer[20]; int count = 0; while (cin.getline(buffer,20)) name[count] = buffer; --count; tout << "The names are:\n"; for (int i = 0; i < count; i++) tout << "\t" << i CC ". [" << name[i] << "1" << endl;
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This does not work because the assignment name[count] = buffer; assigns the same pointer to each of the strings name [ 0 ] , name [ 11, etc. Arrays cannot be assigned this way. To copy one array into another, use s trcpy ( ) , or s trncpy ( ) .
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7.16 Write the s trcpy ( ) function.
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This copies the string s 2 into the string s 1: char* strcpy(char* sl, const char* s2)
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*s2; )
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for (char* p = sl; *p++ = *s2++; *p = \O ; return sl;
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STRINGS
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[CHAP. 7
The pointer p is initialized at the beginning of s 1. On each iteration of the for loop, the character * s 2 is copied into the character *p, and then both s 2 and p are incremented. The loop continues until * s 2 is 0 (i.e., the null character I\ 0 I). Then th e null character is appended to the string s 1 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 s 1 should already have been defined to be a character string with the same length as s 2. 7.17
Write the s tmcat
( > function.
This function appends up to n characters from s2 onto the end of s 1. It is the same as the s t r c a t ( ) function except that its third argument n limits the number of characters copied: char* strncat(char* sl, const char* s2, size-t n) -t // find end of sl for (char* end = sl; *end; end++) ; for (char* p = s2; *p && p - s2 c n; ) *end++ = *p++; *end = '\O'; return sl;
The first for loop finds the end of string s 1. That is where the characters from string s 2 are to be appended. The second for loop copies characters from s 2 to the locations that follow s 1. Notice how the extra condition q - s2 c n limits the number of characters copied to n: the expression q s2 equals the number of characters copied because it is the difference between q (which points to the next character to be copied) and s2 (which points to the beginning of the string). Note that this function does not allocate any new storage. It requires that string s 1 have at least k more bytes allocated, where k is the smaller of n and the length of string s 2. 7.18
Write and test a function that returns the plural form of the singular English word passed to it.
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 -t pluralize(char* s)
int len = strlen(s); // last letter char* p = s + len - 1; // last 2 letters char* q = s + len - 2; if (*p == 'h' &SC (*q == 'c' II *q == 's')) strcat(p, else if (*p == 's') strcat(p, "es"); else if (*p == 'y') if (isvowel( strcat(p, "s"); else strcpy(p, "ies"); else if (*p == 'z') if (isvowel( strcat(p, "zes"); else strcat(p, "es"); else strcat(p, 'Is"); >
"es");
CHAP. 71
STRINGS
Two of the tests depend upon whether the second from last letter is a vowel, so we define a little boolean function i svowel ( ) for testing that condition:
int isvowel(char c) 11
return (c == 'a'
u );
The test driver repeatedly reads a word, prints it, pluralizes it, and prints it again. The loop terminates when the user enters a single blank for a word: #include #include <iostream.h> <string.h>
void pluralize(char*); main0 1 char word[80]; for (;;) 1 cin.getline(word, 80); if (*word == ' ') break; tout << '\tThe singular is [" << word << "].\n"; pluralize(word); tout << "\t The plural is [" << word << "].\n"; ,
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