被接受的答案中第一个(得票较高)评论抱怨现有std集合操作缺少运算符。
一方面,我理解标准库中缺少这些运算符。另一方面,如果需要,很容易添加它们(为了个人的快乐)。
我重载了
operator *()
用于交集
operator +()
用于并集。
示例 test-set-ops.cc
:
#include <algorithm>
#include <iterator>
#include <set>
template <class T, class CMP = std::less<T>, class ALLOC = std::allocator<T> >
std::set<T, CMP, ALLOC> operator * (
const std::set<T, CMP, ALLOC> &s1, const std::set<T, CMP, ALLOC> &s2)
{
std::set<T, CMP, ALLOC> s;
std::set_intersection(s1.begin(), s1.end(), s2.begin(), s2.end(),
std::inserter(s, s.begin()));
return s;
}
template <class T, class CMP = std::less<T>, class ALLOC = std::allocator<T> >
std::set<T, CMP, ALLOC> operator + (
const std::set<T, CMP, ALLOC> &s1, const std::set<T, CMP, ALLOC> &s2)
{
std::set<T, CMP, ALLOC> s;
std::set_union(s1.begin(), s1.end(), s2.begin(), s2.end(),
std::inserter(s, s.begin()));
return s;
}
#include <iostream>
using namespace std;
template <class T>
ostream& operator << (ostream &out, const set<T> &values)
{
const char *sep = " ";
for (const T &value : values) {
out << sep << value; sep = ", ";
}
return out;
}
int main()
{
set<int> s1 { 1, 2, 3, 4 };
cout << "s1: {" << s1 << " }" << endl;
set<int> s2 { 0, 1, 3, 6 };
cout << "s2: {" << s2 << " }" << endl;
cout << "I: {" << s1 * s2 << " }" << endl;
cout << "U: {" << s1 + s2 << " }" << endl;
return 0;
}
已编译并测试:
$ g++ -std=c++11 -o test-set-ops test-set-ops.cc
$ ./test-set-ops
s1: { 1, 2, 3, 4 }
s2: { 0, 1, 3, 6 }
I: { 1, 3 }
U: { 0, 1, 2, 3, 4, 6 }
$
我不喜欢运算符中返回值的复制。也许可以使用移动赋值来解决,但这仍然超出了我的能力范围。
由于我对这些“新奇”的移动语义知识有限,我担心运算符返回可能会导致返回集合的副本。
Olaf Dietsche指出这种担忧是不必要的,因为
std::set
已经配备了移动构造/赋值。
尽管我相信他,但我在考虑如何检查这一点(类似于"自我说服")。实际上,这很容易。由于模板必须在源代码中提供,因此您可以使用调试器逐步进行排查。因此,我在
operator *()
的
return s;
处设置了断点,并进行了单步操作,立即进入了
std::set::set(_myt&& _Right)
:et voilà——移动构造函数。感谢Olaf给我的启示。
为了完整起见,我也实现了相应的赋值运算符。
operator *=()
用于集合的“破坏性”交集
operator +=()
用于集合的“破坏性”并集。
示例test-set-assign-ops.cc
:
#include <iterator>
#include <set>
template <class T, class CMP = std::less<T>, class ALLOC = std::allocator<T> >
std::set<T, CMP, ALLOC>& operator *= (
std::set<T, CMP, ALLOC> &s1, const std::set<T, CMP, ALLOC> &s2)
{
auto iter1 = s1.begin();
for (auto iter2 = s2.begin(); iter1 != s1.end() && iter2 != s2.end();) {
if (*iter1 < *iter2) iter1 = s1.erase(iter1);
else {
if (!(*iter2 < *iter1)) ++iter1;
++iter2;
}
}
while (iter1 != s1.end()) iter1 = s1.erase(iter1);
return s1;
}
template <class T, class CMP = std::less<T>, class ALLOC = std::allocator<T> >
std::set<T, CMP, ALLOC>& operator += (
std::set<T, CMP, ALLOC> &s1, const std::set<T, CMP, ALLOC> &s2)
{
s1.insert(s2.begin(), s2.end());
return s1;
}
#include <iostream>
using namespace std;
template <class T>
ostream& operator << (ostream &out, const set<T> &values)
{
const char *sep = " ";
for (const T &value : values) {
out << sep << value; sep = ", ";
}
return out;
}
int main()
{
set<int> s1 { 1, 2, 3, 4 };
cout << "s1: {" << s1 << " }" << endl;
set<int> s2 { 0, 1, 3, 6 };
cout << "s2: {" << s2 << " }" << endl;
set<int> s1I = s1;
s1I *= s2;
cout << "s1I: {" << s1I << " }" << endl;
set<int> s2I = s2;
s2I *= s1;
cout << "s2I: {" << s2I << " }" << endl;
set<int> s1U = s1;
s1U += s2;
cout << "s1U: {" << s1U << " }" << endl;
set<int> s2U = s2;
s2U += s1;
cout << "s2U: {" << s2U << " }" << endl;
return 0;
}
编译并测试:
$ g++ -std=c++11 -o test-set-assign-ops test-set-assign-ops.cc
$ ./test-set-assign-ops
s1: { 1, 2, 3, 4 }
s2: { 0, 1, 3, 6 }
s1I: { 1, 3 }
s2I: { 1, 3 }
s1U: { 0, 1, 2, 3, 4, 6 }
s2U: { 0, 1, 2, 3, 4, 6 }
$