无限循环录音字符串

Question

代码应该创建一个双向链表。然后应将 IP 地址列表添加到此列表中，其中包含满足唯一 IP 的次数。然后应该对列表进行排序。对不起，代码在记录时循环到某个地方。以粗体突出显示该位置（尝试这样做:)）。 附言如果您能帮助我选择排序方法，我会很高兴。我已经做了一个，但是使用快速排序或其他什么会更好？

#include <stdlib.h>
#include <iostream>
#include <stdio.h>
using namespace std;
struct IP
{
    char b[20];
    int count;
};
struct Node
{
    IP a;

    Node *Next,*Prev;
};
struct List
{
    Node *Head,*Tail;
    int length;
    List():Head(NULL),Tail(NULL){};


};
List* list_new()
{
    return (List *)calloc(1, sizeof(List));
}
void list_delete(List* l)
{
    while (l->Head)
    {
        l->Tail=l->Head->Next;
        free (l->Head);
        l->Head=l->Tail;
    }
    l->length=0;
}
bool push(List* l, IP a)
{
    Node *temp=(Node* ) calloc (1, sizeof(Node));
    temp->Next=NULL;
    temp->a=a;
    if (l->Head!=NULL)
    {
        temp->Prev=l->Tail;
        l->Tail->Next=temp;
        l->Tail=temp;
    }
    else
    {
        temp->Prev=NULL;
        l->Head=l->Tail=temp;
    }
    return 1;
}
bool pop(List*l, IP* x)
{
    (*x)=l->Tail->a;
    l->Tail->Prev->Next=NULL;
    l->Tail=l->Tail->Prev;
    l->length++;
    return 1;
}
bool unshift(List*l, IP a)
{
    Node *temp=(Node* ) calloc (1, sizeof(Node));
    temp->Next=NULL;
    temp->a=a;
    if (l->Head!=NULL)
    {
        temp->Next=l->Head;
        l->Head->Prev=temp;
        l->Head=temp;


    }
    else
    {
        temp->Prev=NULL;
        l->Head=l->Tail=temp;
    }
    return 1;
}
bool shift(List* l, IP* x)
{
    (*x)=l->Head->a;
    l->Head->Next->Prev=NULL;
    l->Head=l->Head->Next;
    return 1;
}
bool reverse (List* l)
{
    Node* temp=l->Head;
    Node* swaps=NULL;
    l->Tail=l->Head;
    while (temp!=NULL)
    {
        swaps=temp->Prev;
        temp->Prev=temp->Next;
        temp->Next=swaps;
        temp=temp->Prev;
    }
    if (swaps != NULL) l->Head = swaps->Prev;
    return 1;
}
void sort (List* l)
{
    int i;
    for (i=0; i<l->length; ++i) {
        Node* compared = l->Head;
        while (compared->Next != NULL) {
            if (compared->Next->a.count > compared->a.count) {
                IP t = compared->Next->a;
                compared->Next->a = compared->a;
                compared->a = t;
            }
            compared = compared->Next;
        }
    }
}
void Show(List* l)
{
    int i;

    Node* temp=l->Head;
    while (temp!=NULL)
    {

        cout<<temp->a.b<<" "<<temp->a.count<<"\n";
        temp=temp->Next;
    }
    cout<<"\n";
}

int main ()
{
    int i;
    char strbuf[1000],chTemp;
    IP ipTemp;
    bool met;
    system("CLS");

    List* l = list_new();

    FILE* foo;
    errno_t err;
    err=fopen_s(&foo,"input.txt","r");
    if( err == 0 )
    {
        printf( "The file 'input.txt' was opened\n" );
    }
    else
    {
        printf( "The file 'input.txt' was not opened\n" );
    }
    while (!feof(foo))
    {

        fgets(strbuf,1000,foo);
        fclose(foo);
        for (i=0;i++;i<20)
            if (strbuf[i]==' ') {strncpy_s( ipTemp.b,strbuf, i);break;}

        Node* cur = l->Head;
        met=0;
        while (cur!=NULL)
        {
            if (cur->a.b == ipTemp.b)
            {
                met=1;
                cur->a.count++;
                break;
            }
            cur=cur->Next;
        }
        if (met==0)
        {
            push(l,ipTemp);
            l->Tail->a.count++;
        }
    }

    sort(l);
    Show(l);

    system("PAUSE");
}

Answer 1

如果代码有更清晰的缩进，你可能会意识到逻辑是错误的：

while (!feof(foo))
{
    fgets(strbuf,1000,foo);     // <-- what if fgets hits EOF or error occurs?
    fclose(foo);                // <-- why?
    for (i = 0; i++; i < 20)    // <-- i++ is always true ~> infinite loop
        ....
    ...
}

应该是（假设您想用 C 编写代码）：

while (fgets(strbuf, 1000, foo))
{
    for (i = 0; i < 20; i++)
        ....
    ...
}

Answer 2

for (i=0;i++;i<20) 应该是for (i=0;i<20;i++)