template<class InputIterator1, class InputIterator2>
bool
lexicographical_compare(InputIterator1 first1, InputIterator1 last1,
InputIterator2 first2, InputIterator2 last2);
template<class ExecutionPolicy, class ForwardIterator1, class ForwardIterator2>
bool
lexicographical_compare(ExecutionPolicy&& exec,
ForwardIterator1 first1, ForwardIterator1 last1,
ForwardIterator2 first2, ForwardIterator2 last2);
template<class InputIterator1, class InputIterator2, class Compare>
bool
lexicographical_compare(InputIterator1 first1, InputIterator1 last1,
InputIterator2 first2, InputIterator2 last2,
Compare comp);
template<class ExecutionPolicy, class ForwardIterator1, class ForwardIterator2, class Compare>
bool
lexicographical_compare(ExecutionPolicy&& exec,
ForwardIterator1 first1, ForwardIterator1 last1,
ForwardIterator2 first2, ForwardIterator2 last2,
Compare comp);
Returns: true if the sequence of elements defined by the range [first1, last1) is lexicographically less than the sequence of elements defined by the range [first2, last2) and false otherwise.
Complexity: At most 2min(last1 - first1, last2 - first2) applications of the corresponding comparison.
Remarks: If two sequences have the same number of elements and their corresponding elements (if any) are equivalent, then neither sequence is lexicographically less than the other. If one sequence is a prefix of the other, then the shorter sequence is lexicographically less than the longer sequence. Otherwise, the lexicographical comparison of the sequences yields the same result as the comparison of the first corresponding pair of elements that are not equivalent.
[ Example: The following sample implementation satisfies these requirements:
for ( ; first1 != last1 && first2 != last2 ; ++first1, (void) ++first2) {
if (*first1 < *first2) return true;
if (*first2 < *first1) return false;
}
return first1 == last1 && first2 != last2;— end example ]