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Trigraphs were proposed for deprecation in C++0x, which was released as C++11. [13] This was opposed by IBM, speaking on behalf of itself and other users of C++, [14] and as a result trigraphs were retained in C++11. Trigraphs were then proposed again for removal (not only deprecation) in C++17. [15]
For example, one might wish to find all occurrences of a "word" despite it having alternate spellings, prefixes or suffixes, etc. Another more complex type of search is regular expression searching, where the user constructs a pattern of characters or other symbols, and any match to the pattern should fulfill the search.
The erase–remove idiom cannot be used for containers that return const_iterator (e.g.: set) [6] std::remove and/or std::remove_if do not maintain elements that are removed (unlike std::partition, std::stable_partition). Thus, erase–remove can only be used with containers holding elements with full value semantics without incurring resource ...
In object-oriented languages, string functions are often implemented as properties and methods of string objects. In functional and list-based languages a string is represented as a list (of character codes), therefore all list-manipulation procedures could be considered string functions.
a modified_identifier_list is a comma-separated list of two or more occurrences of modified_identifier; and a declarator_list is a comma-separated list of declarators, which can be of the form identifier As object_creation_expression (object initializer declarator) ,
Map all occurrences of a, e, i, o, u, y, h, w. to zero(0) Replace all consonants (include the first letter) with digits as in [2.] above. Replace all adjacent same digits with one digit, and then remove all the zero (0) digits; If the saved letter's digit is the same as the resulting first digit, remove the digit (keep the letter).
For example, String.class can be used instead of doing new String().getClass(). continue Used to resume program execution at the end of the current loop body. If followed by a label, continue resumes execution at the end of the enclosing labeled loop body. default
The picture shows two strings where the problem has multiple solutions. Although the substring occurrences always overlap, it is impossible to obtain a longer common substring by "uniting" them. The strings "ABABC", "BABCA" and "ABCBA" have only one longest common substring, viz. "ABC" of length 3.