C++ 如何比较两个 std::set？

how to compare two std::set?

提问by Yuushi

I do such comparison of two std::set

我做了两个这样的比较 std::set

#include <cstdlib>
#include <cstdio>
using namespace std;

#include <vector>
#include <set>


int main(int argc, char** argv)
{
    int myints1[]= {10,20,30,40,50};
    int myints2[]= {50,40,30,20,10};
    std::set<int> s1 (myints1,myints1+5);
    std::set<int> s2(myints2,myints2+5);
    if(s1==s2){
        printf("sets: true");
    }else printf("sets: false");
    std::set<int>::iterator it2=s2.begin();
    for(std::set<int>::iterator it1=s1.begin();it1!=s1.end();it1++){
                printf("\ns1: %d  s2: %d",*it1,*it2);
        it2++;
    }
}

output:

输出：

sets: true
s1: 10  s2: 10
s1: 20  s2: 20
s1: 30  s2: 30
s1: 40  s2: 40
s1: 50  s2: 50

Question:

题：

Is this the right way to do it? Or is any other (special) way of comparing two sets?

这是正确的方法吗？或者是比较两组的任何其他（特殊）方式？

回答by Yuushi

Yes, operator==is correctly defined for all standard containers (exceptthe unordered containers - based on 23.2.5.2 of the standard), and will generally do a lexicographic comparison. See for example here. The relevant quote:

是的，operator==已为所有标准容器（无序容器除外- 基于标准的 23.2.5.2）正确定义，并且通常会进行字典比较。参见此处的示例。相关报价：

Checks if the contents of lhs and rhs are equal, that is, whether lhs.size() == rhs.size() and each element in lhs has equivalent element in rhs at the same position.

检查 lhs 和 rhs 的内容是否相等，即 lhs.size() == rhs.size() 和 lhs 中的每个元素是否在相同位置具有 rhs 中的等效元素。

Since std::setis an ordered container, any set with the same size and same elements (given the comparators are the same) will necessarily have them in the same position, hence will compare equal.

由于std::set是有序容器，任何具有相同大小和相同元素的集合（假设比较器相同）将必然将它们放在相同的位置，因此比较相等。

回答by WiSaGaN

There are several set operations in C++ standard library header <algorithm>.

C++ 标准库头文件中有几个集合操作<algorithm>。

std::set_differencegives those elements that are in set 1 but not set 2.

std::set_difference给出那些在集合 1 但不在集合 2 中的元素。

std::set_intersectiongives those elements that are in both sets.

std::set_intersection给出在两个集合中的那些元素。

std::set_symmetric_differencegives those elements that appear in one of the sets but not both.

std::set_symmetric_difference给出出现在其中一个集合中但不出现在两个集合中的那些元素。

std::set_uniongives those elements that are in either set 1 or set 2.

std::set_union给出那些在集合 1 或集合 2 中的元素。

The algorithms above can also be applied to STL containers other than std::set, but the containers have to be sorted first (std::setis sorted by default).

上述算法也可以应用于 STL 容器以外的 STL 容器std::set，但容器必须先排序（std::set默认排序）。

回答by Rishiraj Surti

Another way would be this:

另一种方法是这样的：

template<typename Set>

bool set_compare(Set const &lhs, Set const &rhs){
    return lhs.size() == rhs.size() 
        && equal(lhs.begin(), lhs.end(), rhs.begin());
}

Inspired from the elegant answer here.

灵感来自这里优雅的答案。