c++ - 归并排序的实现

Question

我正在对列表编写合并排序(不要看函数，我仍然构建它们)。问题如下:编译器无法读取 merge_sort 的归纳，我不知道为什么。我想请你帮忙。下面显示了代码。

#include <iostream>
#include <cstdlib>
#include <iomanip>
#include <time.h>
#include <cmath>
using namespace std;

struct mode
{
    int vol;
    mode *next;
};
void push(mode *&head, int x)
{
    mode *pon = new mode;
    pon->vol = x;
    pon->next = head;
    head = pon;
}
void show(mode *head)
{
    cout << "head-> ";
    mode *p = head;

    while (p != NULL)
    {
        cout << p->vol << " -> ";
        p = p->next;
    }
    cout << "NULL" << endl;
}

void split_list(mode*& p, mode*& p2, mode*& p3)
{
    int middle, counter = 0;
    mode* p_head = p; // p_head, p -> 4,5,6,2,2,1

    while (p->next)
    {
        counter = counter + 1; // 6
        p = p->next;
    }

    counter = counter + 1;
    middle = counter / 2; // 3

    counter = 0;
    p = p_head;
    p3 = p;

    while (counter != middle) // 0!= 3 p,p_head -> 4, p3 -> 5 ||  1!=3 p,p_head ->4->5->p3->6 || 2!=3 p,p_head ->4->5->6 ->p3>2 || ~(3!=3)
    {
        p3 = p3->next;
        counter = counter + 1; // p, p_head -> 4,5,6, p3-> 2,2,1
    }

    p2 = p_head; // p2,p,p_head -> 4,5,6 p3-> 2,2,1

    while (p2->next != p3) // 5!=2 || 6!=2 || ~(2!=2) => p,p_head ->4,5,p2->6=NULL p3->2->2->1
    {
        p2 = p2->next;
    }

    p2->next = NULL; // 4,5,6 p2 -> null , p3 -> 2,2,1
    p2 = p_head;

    //return &*p2,*p3;
    delete p,p_head;

}

void merge_sort (mode*& head)
{
    mode* p = head;
    mode *p2 = NULL, *p3 = NULL;

    if (head && head->next)
    {
        split_list(p, p2, p3);
        cout << p2->vol << endl;
    }
}

int main()
{
    mode* head = NULL;
    mode* head2 = NULL;
    mode* head3 = NULL;

    push(head, 5);
    push(head, 103);
    push(head, 100);
    push(head, 12);
    push(head, 1052);
    push(head, 10);


    show(head);

    merge_sort(head) << endl;
    //show(head);

    system("pause");
}

Answer 1

在你的main功能，我发现以下行:

merge_sort(head) << endl;

它试图调用方法 operator<<关于 merge_sort 的返回值这是不可能的，因为merge_sort是 void .

我认为这会按预期运行:

std::cout << merge_sort(head) << std::endl;

如你所见，我使用了 std:: . StackOverflow 上的这个问题(及其接受的答案)是为什么:Why is "using namespace std" considered bad practice?