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Why couldn t the second constructor for our String class have a default value for its first argument, like this:
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String(char c=' I, unsigned n=O)
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What is wrong with using the copy constructor that is automatically provided by the compiler instead of writing our own copy constructor explicitly What is wrong with using the assignment operator that is automatically provided by the compiler instead of writing our own assignment operator explicitly In what ways is our String class more efficient than simply using C-strings In what ways is it less efficient.
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Implement the String comparison operator = = directly, without using functions from the standard s tr ing . h header file.
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We have the same function header. But now we have to check the object s data members directly: int operator== (const String& sl, const String& s2)
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if (sl.len != s2.len) return 0; for (int i = 0; i < sl.len; i++) if (sl .buf[i] != s2.buf[i]) return 0; return 1;
Since we are storing the string lengths, we can determine immediately that the two strings are not equal if their len fields are not the same. Otherwise, we scan through the two strings in parallel, comparing corresponding characters. If a single mismatch is found, we can return 0 immediately. Only if all the corresponding characters match can we conclude that the two strings are equal and return 1.
Implement and test the following member function for the string class:
istream& getline(istream& istr, char c='\n');
This function reads a line of characters from the input stream object i s t r until it encounters the character C. These characters are stored in the object s buffer, and the input stream object is returned.
We have the same function header. But now we have to check the object s data members directly: istream& String: :getline(istream& 1 char temp[256]; istr.getline(temp, 256, c); len = strlen(temp); delete [] buf; buf = new char[len + 11; strcpy(buf, temp); return istr; > istr, char c='\n')
As with the overloaded extraction operator >>, this function uses a temporary C-string buffer of 256 characters. It invokes the get 1 ine ( ) functions defined in < ios tream . h> to read the line. Then it performs the same steps that are used in the third constructor to transform the C-string buffer temp into the string object. Here is a test driver for this function: #include "String.h"
main0 -t String s; s.getline(cin); tout << "\t[" << s << "]\n"; 'I '>; s.getline(cin, tout << "\t[" << s << "]\n"; s.getline(cin, 'I '>; tout << "\t[" << s << "]\n";
A String CLASS
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Here is the output:
The first call uses the default value \ n I for the delimiter argument c, so it reads the entire line. The following two calls use the character I I 1 for the delimiter, each reads only up to the next occurrence of that character. The effect is to be able to use the delimiter as a separator between input fields.
Implement and test the following member function for the String class:
int firstLocation(const String& s, unsigned k=O);
This searches the object s buffer, beginning with character buf [ k] for the string S. If s is found to be a substring, then the index of its first occurrence is returned; otherwise -1 is returned.
In this solution, we implement a brute force searching method. Improvements could be made by using more efficient pattern-matching algorithms, such as the Knuth-Morris-Pratt Algorithm, the BoyerMoore Algorithm, or the Rabin-Karp Algorithm. (See 19 in [Savitch].) int String: :firstLocation(const String& s, unsigned k=O)
for (int i = k, j = 0; i c len &SC j c s.len; if+, j++) if (buf[i] != s.buf[j]) { i -= j; j = -1; 1 if (j == s.len) return i - s.len; // substring found else return -1;
In this implementation, the for loop compares buf [ i ] with s . buf [ j ] , incrementing i and j simultaneously, and resetting i and j whenever a mismatch is found. For example, consider the call: x.firstLocation(z). When i =5and j =0, buf [i] matches s.buf [j];theyareboth 'F'. So i and j both increment to i = 6 and j =l,andagain buf [i] matches s.buf [j]; this time they are both G . So i and j both increment to i = 7 and j = 2. But this time they do not match: buf[i] = 'H' and s.buf[i] = 'Z'. So i is reset to 5, and j is reset to -1. But then they both increment again before the next comparison is made, so next bu f [ 6 ] is compared with s. buf [ 01. They don t match, so next buf [ 71 is compared with s. buf [ 01. The loop terminates when either i = len or j = s . 1 en. If ( j = = s . 1 en) , then the substring was found, because buf [ i ] matched s . buf [ j ] for each j from 0 to s . 1 en-l. In this case, i is pointing to the character immediately after the last character in the match, so i - s . len will point to the first character in the match and that is the location in buf that should be returned.
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