c++ 为什么gquiz1的上限值小于键值

m3eecexj  于 2022-12-01  发布在  其他
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// CPP Program to demonstrate the
// implementation of multiset
#include <iostream>
#include <iterator>
#include <set>

using namespace std;

int main()
{
    // empty multiset container
    multiset<int, greater<int> > gquiz1;

    // insert elements in random order
    gquiz1.insert(40);
    gquiz1.insert(30);
    gquiz1.insert(60);
    gquiz1.insert(20);
    gquiz1.insert(50);

    // 50 will be added again to
    // the multiset unlike set
    gquiz1.insert(50);
    gquiz1.insert(10);

    // printing multiset gquiz1
    multiset<int, greater<int> >::iterator itr;
    cout << "\nThe multiset gquiz1 is : \n";
    for (itr = gquiz1.begin(); itr != gquiz1.end(); ++itr) {
        cout << *itr << " ";
    }
    cout << endl;

    // assigning the elements from gquiz1 to gquiz2
    multiset<int> gquiz2(gquiz1.begin(), gquiz1.end());

    // print all elements of the multiset gquiz2
    cout << "\nThe multiset gquiz2 \n"
            "after assign from gquiz1 is : \n";
    for (itr = gquiz2.begin(); itr != gquiz2.end(); ++itr) {
        cout << *itr << " ";
    }
    cout << endl;

    // remove all elements up to element
    // with value 30 in gquiz2
    cout << "\ngquiz2 after removal \n"
            "of elements less than 30 : \n";
    gquiz2.erase(gquiz2.begin(), gquiz2.find(30));
    for (itr = gquiz2.begin(); itr != gquiz2.end(); ++itr) {
        cout << *itr << " ";
    }

    // remove all elements with value 50 in gquiz2
    int num;
    num = gquiz2.erase(50);
    cout << "\ngquiz2.erase(50) : \n";
    cout << num << " removed \n";
    for (itr = gquiz2.begin(); itr != gquiz2.end(); ++itr) {
        cout << *itr << " ";
    }

    cout << endl;

    // lower bound and upper bound for multiset gquiz1
    cout << "\ngquiz1.lower_bound(40) : \n"
        << *gquiz1.lower_bound(40) << endl;
    cout << "gquiz1.upper_bound(40) : \n"
        << *gquiz1.upper_bound(40) << endl;

    // lower bound and upper bound for multiset gquiz2
    cout << "gquiz2.lower_bound(40) : \n"
        << *gquiz2.lower_bound(40) << endl;
    cout << "gquiz2.upper_bound(40) : \n"
        << *gquiz2.upper_bound(40) << endl;

    return 0;
}

gquiz1的上限打印30,关键字为40。gquiz2的上限似乎只给出了一个更高的值。我认为这将与多集gquiz1中的多个类似元素有关,但在所有擦除函数后,在gquiz2中插入50两次,gquiz2的上限仍然给出了比关键字更高的值,而gquiz1给出了更低的值。请帮助。
修改了两次插入50的代码。我尝试了两次插入50的代码:

// CPP Program to demonstrate the
// implementation of multiset
#include <iostream>
#include <iterator>
#include <set>

using namespace std;

int main()
{
    // empty multiset container
    multiset<int, greater<int> > gquiz1;

    // insert elements in random order
    gquiz1.insert(40);
    gquiz1.insert(30);
    gquiz1.insert(60);
    gquiz1.insert(20);
    gquiz1.insert(50);

    // 50 will be added again to
    // the multiset unlike set
    gquiz1.insert(50);
    gquiz1.insert(10);

    // printing multiset gquiz1
    multiset<int, greater<int> >::iterator itr;
    cout << "\nThe multiset gquiz1 is : \n";
    for (itr = gquiz1.begin(); itr != gquiz1.end(); ++itr) {
        cout << *itr << " ";
    }
    cout << endl;

    // assigning the elements from gquiz1 to gquiz2
    multiset<int> gquiz2(gquiz1.begin(), gquiz1.end());

    // print all elements of the multiset gquiz2
    cout << "\nThe multiset gquiz2 \n"
            "after assign from gquiz1 is : \n";
    for (itr = gquiz2.begin(); itr != gquiz2.end(); ++itr) {
        cout << *itr << " ";
    }
    cout << endl;

    // remove all elements up to element
    // with value 30 in gquiz2
    cout << "\ngquiz2 after removal \n"
            "of elements less than 30 : \n";
    gquiz2.erase(gquiz2.begin(), gquiz2.find(30));
    for (itr = gquiz2.begin(); itr != gquiz2.end(); ++itr) {
        cout << *itr << " ";
    }

    // remove all elements with value 50 in gquiz2
    int num;
    num = gquiz2.erase(50);
    cout << "\ngquiz2.erase(50) : \n";
    cout << num << " removed \n";
    for (itr = gquiz2.begin(); itr != gquiz2.end(); ++itr) {
        cout << *itr << " ";
    }

   gquiz2.insert(50);
   gquiz2.insert(50);

    cout << endl;

    // lower bound and upper bound for multiset gquiz1
    cout << "\ngquiz1.lower_bound(40) : \n"
        << *gquiz1.lower_bound(40) << endl;
    cout << "gquiz1.upper_bound(40) : \n"
        << *gquiz1.upper_bound(40) << endl;

    // lower bound and upper bound for multiset gquiz2
    cout << "gquiz2.lower_bound(40) : \n"
        << *gquiz2.lower_bound(40) << endl;
    cout << "gquiz2.upper_bound(40) : \n"
        << *gquiz2.upper_bound(40) << endl;

    return 0;
}

除外:

The multiset gquiz1 is : 
60 50 50 40 30 20 10 

The multiset gquiz2 
after assign from gquiz1 is : 
10 20 30 40 50 50 60 

gquiz2 after removal 
of elements less than 30 : 
30 40 50 50 60 
gquiz2.erase(50) : 
2 removed 
30 40 60 

gquiz1.lower_bound(40) : 
40
gquiz1.upper_bound(40) : 
30
gquiz2.lower_bound(40) : 
40
gquiz2.upper_bound(40) : 
30

实际值:

The multiset gquiz1 is : 
60 50 50 40 30 20 10 

The multiset gquiz2 
after assign from gquiz1 is : 
10 20 30 40 50 50 60 

gquiz2 after removal 
of elements less than 30 : 
30 40 50 50 60 
gquiz2.erase(50) : 
2 removed 
30 40 60 

gquiz1.lower_bound(40) : 
40
gquiz1.upper_bound(40) : 
30
gquiz2.lower_bound(40) : 
40
gquiz2.upper_bound(40) : 
50
h7wcgrx3

h7wcgrx31#

你的问题就在这里

multiset<int, greater<int> > gquiz1;

因为您使用的是greater<int>,所以upper_boundlower_bound的通常意义已经改变。使用upper_bound将返回值 * 小于 * 键的第一个元素,而lower_bound将返回值 * 大于或等于 * 键的第一个元素。
upper_bound的精确定义是,对于map〈T,C〉和给定的key,它返回第一个C(key,x)为真的元素x。如果C是大于运算,那么这个定义意味着你将得到第一个小于key的元素。换句话说,upper_bound的通常含义是相反的。
如果更改为此

multiset<int> gquiz1;

您将看到upper_boundlower_bound的通常含义返回。

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