because the input ranges may contain duplicate elements. The generalization is that if a value appears m times in [first1, last1) and n times in [first2, last2) (where m or n may be zero), then it appears max(m-n, 0) times in the output range. [1] Set_difference is stable, meaning both that elements are copied from the first range rather than the second, and that the relative order of elements in the output range is the same as in the first input range.
The two versions of set_difference differ in how they define whether one element is less than another. The first version compares objects using operator<, and the second compares objects using a function object comp.
Definition Defined in the standard header algorithm, and in the nonstandard backward-compatibility header algo.h.
Requirements on types For the first version:
• InputIterator1 is a model of Input Iterator.
• InputIterator2 is a model of Input Iterator.
• OutputIterator is a model of Output Iterator.
• InputIterator1 and InputIterator2 have the same value type.
• InputIterator's value type is a model of LessThan Comparable.
• The ordering on objects of InputIterator1's value type is a strict weak ordering, as defined in the LessThan Comparable requirements.
• InputIterator's value type is convertible to a type in OutputIterator's set of value types.
For the second version:
• InputIterator1 is a model of Input Iterator.
• InputIterator2 is a model of Input Iterator.
• OutputIterator is a model of Output Iterator.
• StrictWeakOrdering is a model of Strict Weak Ordering.
• InputIterator1 and InputIterator2 have the same value type.
• InputIterator1's value type is convertible to StrictWeakOrdering's argument type.
• InputIterator's value type is convertible to a type in OutputIterator's set of value types.
Preconditions For the first version:
• [first1, last1) is a valid range.
• [first2, last2) is a valid range.
• [first1, last1) is ordered in ascending order according to operator<. That is, for every pair of iterators i and j in [first1, last1) such that i precedes j, *j < *i is false.
• [first2, last2) is ordered in ascending order according to operator<. That is, for every pair of iterators i and j in [first2, last2) such that i precedes j, *j < *i is false.
• There is enough space to hold all of the elements being copied. More formally, the requirement is that [result, result + n) is a valid range, where n is the number of elements in the union of the two input ranges.
• [first1, last1) and [result, result + n) do not overlap.
• [first2, last2) and [result, result + n) do not overlap.
For the second version:
• [first1, last1) is a valid range.
• [first2, last2) is a valid range.
• [first1, last1) is ordered in ascending order according to comp. That is, for every pair of iterators i and j in [first1, last1) such that i precedes j, comp(*j, *i) is false.
• [first2, last2) is ordered in ascending order according to comp. That is, for every pair of iterators i and j in [first2, last2) such that i precedes j, comp(*j, *i) is false.
• There is enough space to hold all of the elements being copied. More formally, the requirement is that [result, result + n) is a valid range, where n is the number of elements in the union of the two input ranges.
• [first1, last1) and [result, result + n) do not overlap.
• [first2, last2) and [result, result + n) do not overlap.
Complexity Linear. Zero comparisons if either [first1, last1) or [first2, last2) is empty, otherwise at most 2 * ((last1 – first1) + (last2 – first2)) – 1 comparisons.
Example inline bool lt_nocase(char c1, char c2) { return tolower(c1) < tolower(c2); }
int main() {
int A1[] = {1, 3, 5, 7, 9, 11};
int A2[] = {1, 1, 2, 3, 5, 8, 13};
char A3[] = {'a', 'b', 'b', 'B', 'B', 'f', 'g', 'h', 'H'};
char A4[] = {'A', 'B', 'B', 'C', 'D', 'F', 'F', 'H' };
const int N1 = sizeof(A1) / sizeof(int);
const int N2 = sizeof(A2) / sizeof(int);
const int N3 = sizeof(A3);
const int N4 = sizeof(A4);
cout << 'Difference of A1 and A2: ';
set_difference(A1, A1 + N1, A2, A2 + N2, ostream_iterator<int>(cout, ' '));
cout << endl << 'Difference of A3 and A4: ';
set_difference(A3, A3 + N3, A4, A4 + N4, ostream_iterator<char>(cout, ' '), lt_nocase);
cout << endl;
}
The output is
Difference of A1 and A2: 7 9 11
