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Before we add to our set of text query routines, let's briefly cover the remaining search functions supported by the string class In addition to find() and find_first_of(), the string class supports several other find operations: rfind() searches for the last that is, rightmost occurrence of the indicated substring For example:
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string river( "Mississippi" ); string::size_type first_pos = riverfind( "is" ); string::size_type last_pos = riverrfind( "is" );
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find() returns an index of 1, indicating the start of the first "is", while rfind() returns an index of 4, indicating the start of the last occurrence of "is" find_first_not_of() searches for the first character of the string that does not match any element of the search string For example, to find the first non-numeric character of a string, we can write
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string elems( "0123456789" ); string dept_code( "03714p3" ); // returns index to the character 'p' string::size_type pos = dept_codefind_first_not_of(elems);
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find_last_of() searches for the last character of the string that matches any element of the search string find_last_not_of() searches for the last character of the string that does not match any element of the search string Each of these operations takes an optional second argument indicating the position within the string to begin searching Exercise 613 Write a program that, given the string
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"ab2c3d7R4E6"
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finds each numeric character and then each alphabet character first using find_first_of() and then find_first_not_of() Exercise 614 Write a program that, given the string
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line1 = "We were her pride of 10 she named us --"; line2 = "Benjamin, Phoenix, the Prodigal" line3 = "and perspicacious pacific Suzanne"; sentence = line1 + ' ' + line2 + ' ' + line3;
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counts the number of words in the sentence and identifies the largest and smallest words If more than one word is either the largest or
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smallest, keep track of all of them
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Handling Punctuation
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Now that we've separated each line of text into individual words, we need to remove any punctuation that may have stuck to the word For example, the following line
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magical but untamed "Daddy, shush, there is no such thing,"
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separates as follows:
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How can we remove the unwanted punctuation First, we'll define a string with all the punctuation elements we wish to remove:
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string filt_elems( "\",;:! )(\\/" );
(The \" and \\ sequences indicate that the quotation mark in the first sequence and the second backslash in the second sequence are to be treated as literal elements within the quoted string and not as the end of the string or as the continuation character to the next line) Next, we'll use the find_first_of() operation to find each matching element, if any, within our string:
while (( pos = wordfind_first_of( filt_elems, pos )) != string::npos )
Finally, we need to erase() the punctuation character from the string:
worderase(pos,1);
The first argument to this version of the erase() operation indicates the position within the string to begin removing characters An optional second argument indicates the number of characters to delete In our example, we are deleting the one character located at
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file:///F|/WinDDK/resources/CPPPrimer/c++primerhtm
pos If we omit the second argument, erase() removes all the characters from pos to the end of the string Here is the full listing of filter_text() It takes two arguments: a pointer to our string vector containing the text, and a string object containing the elements to filter
void filter_text( vector<string> *words, string filter ) { vector<string>::iterator iter = words->begin(); vector<string>::iterator iter_end = words->end(); // if no filter is provided by user, default to a minimal set if ( ! filtersize() ) filterinsert( 0, "\"," ); while ( iter != iter_end ) { string::size_type pos = 0; // for each element found, erase it while (( pos = (*iter)find_first_of( filter, pos )) != string::npos ) (*iter)erase(pos,1); iter++; } }
Do you see why we do not increment pos with each iteration of the loop That is, do you see why the following is incorrect
while (( pos = (*iter)find_first_of( filter, pos )) != string::npos ) { (*iter)erase(pos,1); ++pos; // not correct }