Replacing string characters

The insert( ) function is particularly nice because it absolves you of making sure the insertion of characters in a string won?t overrun the storage space or overwrite the characters immediately following the insertion point. Space grows, and existing characters politely move over to accommodate the new elements. Sometimes, however, this might not be what you want to happen. If you want the size of the string to remain unchanged, use the replace( ) function to overwrite characters. There are quite a number of overloaded versions of replace( ), but the simplest one takes three arguments: an integer indicating where to start in the string, an integer indicating how many characters to eliminate from the original string, and the replacement string (which can be a different number of characters than the eliminated quantity). Here?s a simple example:

The tag is first inserted into s (notice that the insert happens before the value indicating the insert point and that an extra space was added after tag), and then it is found and replaced.

You should actually check to see if you?ve found anything before you perform a replace( ). The previous example replaces with a char*, but there?s an overloaded version that replaces with a string. Here?s a more complete demonstration replace( );

If replace doesn?t find the search string, it returns string::npos. The npos data member is a static constant member of the string class that represents a nonexistent character position.

Unlike insert( ), replace( ) won?t grow the string?s storage space if you copy new characters into the middle of an existing series of array elements. However, it will grow the storage space if needed, for example, when you make a ?replacement? that would expand the original string beyond the end of the current allocation. Here?s an example:

The call to replace( ) begins ?replacing? beyond the end of the existing array, which is equivalent to an append operation. Notice that in this example replace( ) expands the array accordingly.

You can easily write such a function using the find( ) and replace( ) member functions as follows:

The version of find( ) used here takes as a second argument the position to start looking in and returns string::npos if it doesn?t find it. It is important to advance the position held in the variable lookHere past the replacement string, of course, in case from is a substring of to. The following program tests the replaceAll function:

As you can see, the string class by itself doesn?t solve all possible problems. Many solutions have been left to the algorithms in the Standard library, because the string class can look just like an STL sequence (by virtue of the iterators discussed earlier). All the generic algorithms work on a ?range? of elements within a container. Usually that range is just ?from the beginning of the container to the end.? A string object looks like a container of characters: to get the beginning of the range you use string::begin( ), and to get the end of the range you use string::end( ). The following example shows the use of the replace( ) algorithm to replace all the instances of the single character ?X? with ?Y?:

Notice that this replace( ) is not called as a member function of string. Also, unlike the string::replace( ) functions that only perform one replacement, the replace( ) algorithm replaces all instances of one character with another.

The replace( ) algorithm only works with single objects (in this case, char objects) and will not replace quoted char arrays or string objects. Since a string behaves like an STL sequence, a number of other algorithms can be applied to it, which might solve other problems that are not directly addressed by the string member functions.

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