0

txt 文件中的值采用以下格式： 

4  
2 3  
5 6  
3 7  
6 9

并且输出必须如下所示：

[2:3][5:6][3:7][6:9]

这是我的代码：

#include <iostream>  
#include <stdio.h>  
using namespace std;  

class node {
    public:  
        int info;
        node* next;
        node(){
            next = NULL;
        }
        node (int value){
            info = value;
            next = NULL;
        }
};

class list {
    private:
        node* head;
    public: 
        list() {            //Constructor
            head = NULL;
        }

    void insert(int value){
        if (head == NULL){
            head = new node;
            head -> info = value;
            return;
        }
        node *temp = head;
        while (temp) {
            temp = temp -> next;
        }
        temp -> next = new node;
        temp -> info = value;
    }

    void showlist(){
        node* temp = head;
        temp = temp -> next;        //ignore first number in txt file
        cout << "Liste \n" << endl;
        while (temp){
            printf ("%d", temp -> info);
            //cout << temp -> info << endl;
            temp = temp -> next;
        }
    }   

    ~list() {           //Destructor
        node* temp = head;
        while (head -> next != NULL) {
            delete temp -> next;
            temp -> next = NULL;
            temp = temp -> next;
        }
        delete head -> next;
        head -> next = NULL;
    }
};

int main(int argc, const char * argv[]) {
    list stone;

    FILE* fp;

    if (argc > 1)
        fp = fopen(argv[1], "r");
    else 
        fp = fopen("output.txt", "r");

    if (!fp)
        printf("Can't open file \n");
    else {
        int value, state;
        int i = 0;
        do {
            state = fscanf(fp, "%d", &value);
            if (state != EOF){
                stone.insert(value);
            }
        stone.showlist();
        }
        while (state != EOF);
        fclose (fp);
    }

    return 0;
}

没有错误，但如果我想执行它，我会得到一个崩溃报告。我忽略了请求的格式作为开始。

4

3 回答 3

1

您的代码中有几个错误。我将列出几个明显的供您考虑：

  1. 你怎么处理头？您在显示列表时忽略了头部，但它实际上包含第一个值。
  2. 每次尝试使用时node->next，都应该保证节点不为NULL
  3. 正如第一个答案指出的那样，while循环后的 temp 为 NULL。解决方案可能是while (temp->next)
  4. 列表的析构函数不正确，您将遇到空列表的分段错误
  5. 您使用printfandfgets而不是iostream，并且使用两个类。你更喜欢哪一个，C 还是 C++？选择一个你喜欢的。
于 2013-06-02T15:52:42.663 回答
0

这个块insert看起来不正确：

    while (temp) {
        temp = temp -> next;
    }
    temp -> next = new node;

在while循环之后，tempis NULL，所以你不能调用temp -> next.

于 2013-06-02T15:23:40.280 回答
0

您似乎想要一个格式化的答案。但是您没有安排将其打印为

[2:3][5:6][3:7][6:9] 

//THE MODIFICATIONS I MADE ARE MINUTE. BUT THEY'LL GIVE YOU THE RESULT YOU DESIRE.
// MODIFY TEXT FILE FOR ANY MORE INPUTS IF YOU NEED TO INSERT MORE
 #include <iostream>  
 #include <stdio.h>  
 #include<conio.h>
 using namespace std;  

class node {
public:  
    int info;
    node* next;
    node(){
        next = NULL;
    }
    node (int value){
        info = value;
        next = NULL;
    }
};

class list {
private:
    node* head;
public: 
    list() {            //Constructor
        head = NULL;
    }

void insert(int value){
    if (head == NULL){
        head = new node;
        head -> info = value;
        return;
    }
    node *temp = head;
    while (temp->next) { //MODIFIED INSERT LOOP TERMINATION CONDN
        temp = temp -> next;
    }
    temp -> next = new node;
    temp -> info = value;
}

void showlist(){
    node* temp = head;
  //  temp = temp -> next;  //ignore first number in txt file _ UNNECESSARY - COMMENTED
    cout << "List \n" << endl;
    while (temp){
        if(!temp->next){break;} //, temp -> info,(temp -> next)-> info);
        else {printf ("[%d:%d]\n", temp -> info,(temp -> next)-> info);}
        //cout << temp -> info << endl;
        temp = (temp -> next)->next;
    }
}   

~list() {           //Destructor
    node* temp = head;
    while (head -> next != NULL) {
        delete temp -> next;
        temp -> next = NULL;
        temp = temp -> next;
    }
    delete head -> next;
    head -> next = NULL;
}
};

int main(int argc, const char * argv[]) {
list stone;

FILE* fp;

if (argc > 1)
    fp = fopen(argv[1], "r");
else 
    fp = fopen("inp.txt", "r"); //CHANGED NAME OF FILE 

if (!fp)
    printf("Can't open file \n");
else {
    int value, state;
    int i = 0;
    do {
        state = fscanf(fp, "%d", &value);
        if (state != EOF){
            stone.insert(value);
        }

    } while (state != EOF);
     stone.showlist();
    fclose (fp);
}
getch();
return 0;
}
于 2013-06-02T16:25:19.350 回答