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while (p) { for ( int i = 0 ; p[i] ; i++ ) if (p[i] == ch) // ch found in current word { count++ ; // referenced by p break ; // finished with current word } // end if (p[i] == ch) p = strtok(NULL, "\t\n \v\f\r" ) ; // advance to next word } // end while (p) return count ; // } 8.20 8.21 void capitalize(char* s) { if (s == NULL) return; for (char* p=s; *p; p++) if (*p>='a' && *p<='z')*p = (char)(*p - 'a' + 'A'); } void removeBlanks( char* s) { if ( s == NULL ) return ; int j = 0 ; for ( int i = 0; s[i] ; i++ ) if ( s[i] != ' ' ) s[j++] = s[i] ; s[j] = '\0' ; } int numWords( const char* s) { if ( s == NULL ) return 0 ; int wordCount = 0 ; char * Copy = new char[ strlen(s) ] ; Copy = strcpy( Copy, s ) ; char * p = strtok( Copy, "\n \v\t\f\r" ) ; while ( p ) { char ch0 = p[0]; // check whether first char is letter if ( ((ch0 >= 'a') && (ch0 <= 'z') ) || // lowercase ((ch0 >= 'A') && (ch0 <= 'Z') ) ) // uppercase wordCount++ ; p = strtok( NULL, "\n \v\t\f\r" ) ; } return wordCount ; } char* reverseWords(char* reverseS, const char* s) { if ( (reverseS == NULL) || (s == NULL) ) return NULL; char * Copy = new char[ strlen(s) ] ; Copy = strcpy( Copy , s ) ; char * currentReverse = new char[ strlen(s) ] ; char * revPtr = reverseS ; *revPtr = '\0' ; // reverse starts with no words char * pS; pS = strtok(Copy, " \t" ) ; // words separated by space or tab while ( pS ) { // reverseS = currentWordInS + currentReverse currentReverse = strcpy( currentReverse, revPtr ) ; revPtr = addWords( revPtr, pS, currentReverse ) ; pS = strtok( NULL, " \t" ) ; // advance pS to next word in s
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} // end while (pS) return revPtr ; } char* { addWords( char* leftPLUSright, const char* left, const char* right ) char * both = leftPLUSright ; const char * pLeft = left ; const char * pRight = right ; while ( *pLeft ) *(both++) = *(pLeft++) ; if ( *left && *right ) // both words nonempty *(both++) = ' ' ; // so put space between while ( *pRight ) *(both++) = *(pRight++) ; *both = '\0' ; // terminate new string with null character return leftPLUSright ;
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9.1 INTRODUCTION The classic C-strings described in 8 are an important part of C++. They provide a very efficient means for fast data processing. But as with ordinary arrays, the efficiency of C-strings comes at a price: the risk of run-time errors, resulting primarily from their dependency upon the use of the NUL character as a string terminator. Standard C++ strings provide a safe alternative to C-strings. By encapsulating the length of the string with the string itself, there is no direct reliance on string terminators. 9.2 FORMATTED INPUT Recall the idea of a stream in C++ as a conduit through which data passes. Input passes through an istream object and output passes through an ostream object. The istream class defines the behavior of objects like cin. The most common behavior is the use of the extraction operator >> (also called the input operator). It has two operands: the istream object from which it is extracting characters, and the object to which it copies the corresponding value formed from those characters. This process of forming a typed value from raw input characters is called formatting. EXAMPLE 9.1 The Extraction Operator >> Performs Formatted Input
Suppose the code cin int n; 4 6 \n cin >> n; istream n int executes on the input 46 This input actually contains the 7 characters: ' ' , ' ', ' ', ' ' , '4', '6', '\n' (four blanks followed by a 4, a 6, and the newline character). It could be viewed as coming through the input stream. The stream object cin scans characters one at a time. If the first character it sees is a whitespace character (a blank, a tab, a newline, etc.), it extracts it and ignores it. It continues to extract and ignore the characters in the stream until it encounters a non-whitespace character. In this example, that would be the '4' . Since the second operand of the expression cin >> n has type int, the cin object is looking for digits to form an integer. So after eating any preceding whitespace, it expects to find one of the 12 characters '+', '-', '0', '1', '2', '3', '4', '5', '6', '7', '8', or '9'. If it encounters any of the other 244 characters, it will fail. In this case, it sees the '4'. So it extracts it and then continues, expecting more digits. As long as it encounters only digits, it continues to extract them. As soon as it sees a non-digit, it stops, leaving that non-digit in the stream. In this case, that means that cin will extract exactly 6 characters: the 4 blanks, the '4', and the '6' . It discards the 4 blanks and then combines the '4' and the '6' to form the integer value 46. Then it copies that value into the object n.
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