链表无限循环

时间:2015-10-21 04:40:30

标签: c algorithm linked-list

尝试创建一个程序来计算在将当前节点移动到头部时使用strcmp进行的比较次数。虽然算法存在一些问题,但它可以使用#34;有时,在其他时候给我一个infiniteloop。现在尝试lldb几个小时,并且每两行代码输入printf消息,但我不知道如何查明问题。我认为它在算法的某个地方,但我看不出它有什么问题。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct Node2
{
    char* word;
    struct Node2 *next;
} Node2;

Node2* head = NULL;
Node2* now = NULL;
int uniqueWords = 0;
int totalMTFRComparisons = 0;


Node2* iNode(char* word)
{   
    Node2* ptr = malloc(sizeof(Node2));
    Node2* tmp = head;
    Node2* prev = head;

    while (tmp)
    {
        totalMTFRComparisons++;
        printf("Current word: %s", tmp->word);
        if (strcmp(tmp->word, word) == 0)
        {
            prev->next = tmp->next;
            tmp->next = head;
            return tmp;
        }
        prev = tmp;
        tmp = tmp->next;
        printf("tmp incremented.\n");
    }

    ptr->word = malloc(strlen(word) + 1);
    strcpy(ptr->word, word);
    ptr->next = NULL;
    if (head == NULL)
    {
        head = now = ptr;
        return ptr;
    }
    else
    {
        ptr->next = head;
        head = ptr;
    }
    return ptr;
}

char* getString()
{
    static char buffer[128];
    while (fgets(buffer, 128, stdin) != NULL)
    {
        iNode(buffer);
    }
    return buffer;
}

int main()
{
    getString();
    printf("Total words: %d, total MTFR comparisons: %d\n", uniqueWords, totalMTFRComparisons);

    Node2* ptr2 = head;
    Node2* tmp2 = head;
    while (ptr2 != NULL)
    {
        tmp2 = ptr2->next;
        free(ptr2->word);
        free(ptr2);
        ptr2 = tmp2;
    }
}

我的控制台只是垃圾邮件: tmp incremented. 但这并不总是发生 - 它只是偶尔发生,所以我不知道造成它的原因。

示例输入&amp;输出:http://pastebin.com/QfuCm7gt

2 个答案:

答案 0 :(得分:3)

你为字符串分配的内存太少了:

ptr->word = malloc(strlen(word));
strcpy(ptr->word, word);

您需要分配strlen(word) + 1个字节以允许空终止符。

当您在分配的空间之外写入时,您将调用未定义的行为。不幸的是,这可能意味着它有时似乎工作正常 - 它是对未定义行为的有效响应,似乎按预期工作,除非它适合系统改变主意并且表现异常。

考虑您是否可以使用strdup()功能。如果没有,请考虑是否应该编写并使用自己的版本。不要忘记检查内存分配是否成功。

(直到你修复了未定义的行为,如果还有其他错误,那么在想知道还有什么问题上真的没有意义。)

我写了这段代码,看看我是否可以模拟你的问题。我不能:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct Node
{
    char *word;
    struct Node *next;
} Node;

int totalMTFRComparisons = 0;

Node* head = NULL;
static Node* iNode(char* word)
{   
    Node* ptr = malloc(sizeof(Node));
    Node* tmp = head;
    Node* prev = head;

    while (tmp)
    {
        totalMTFRComparisons++;
        if (strcmp(tmp->word, word) == 0)
        {
            prev->next = tmp->next;
            tmp->next = head;
            //tmp->next = prev;

/* JL: Still a memory leak here. */
/* Either free(ptr) or don't allocate until after the loop */

            return tmp;
        }
        prev = tmp;
        tmp = tmp->next;
        printf("tmp incremented.\n");
    }
    ptr->word = malloc(strlen(word) + 1);
    strcpy(ptr->word, word);
    ptr->next = NULL;
    if (head == NULL)
    {
        head = ptr;
        return ptr;
    }
    else
    {
        ptr->next = head;
        head = ptr;
    }
    return ptr;
}

static void dump_list(Node *node)
{
    while (node != 0)
    {
        printf("Node word: [[%s]]\n", node->word);
        node = node->next;
    }
}

static void free_list(Node *node)
{
    printf("Freeing list\n");
    while (node != 0)
    {
        Node *next = node->next;
        printf("Freeing: [[%s]]\n", node->word);
        free(node->word);
        free(node);
        node = next;
    }
}

int main(void)
{
    char line[4096];

    while (fgets(line, sizeof(line), stdin) != 0)
    {
        printf("Line: [[%s]]\n", line);
        char *ptr = line;
        char *tok;
        while ((tok = strtok(ptr, "\n\t ")) != 0)
        {
            printf("Word: [[%s]]\n", tok);
            iNode(tok);
            ptr = NULL;
        }
        dump_list(head);
    }
    free_list(head);
    return 0;
}

这或多或少是MCVE。转储和自由函数只允许我确保我可以看到列表中的内容并释放所有内存。

除了修复内存覆盖问题外,我只提供了Node类型的定义,将static放在函数前面(以避免我得到的编译警告之一) ,并添加了两个支持函数和main();我没有改变你的代码。 (顺便说一句,关于第一次编辑的唯一投诉是cur2,我注意到但是忘了删除 - 这非常好;很少有程序,甚至是我编写的程序,都非常干净。)

运行时,我键入:

abc def ghi
mno pqr stuvwxyz humongously-and-tempestuously-but-neither-abstemiously-nor-facetiously-long word!

我得到了输出:

abc def ghi
Line: [[abc def ghi
]]
Word: [[abc]]
Word: [[def]]
tmp incremented.
Word: [[ghi]]
tmp incremented.
tmp incremented.
Node word: [[ghi]]
Node word: [[def]]
Node word: [[abc]]
mno pqr stuvwxyz humongously-and-tempestuously-but-neither-abstemiously-nor-facetiously-long word!
Line: [[mno pqr stuvwxyz humongously-and-tempestuously-but-neither-abstemiously-nor-facetiously-long word!
]]
Word: [[mno]]
tmp incremented.
tmp incremented.
tmp incremented.
Word: [[pqr]]
tmp incremented.
tmp incremented.
tmp incremented.
tmp incremented.
Word: [[stuvwxyz]]
tmp incremented.
tmp incremented.
tmp incremented.
tmp incremented.
tmp incremented.
Word: [[humongously-and-tempestuously-but-neither-abstemiously-nor-facetiously-long]]
tmp incremented.
tmp incremented.
tmp incremented.
tmp incremented.
tmp incremented.
tmp incremented.
Word: [[word!]]
tmp incremented.
tmp incremented.
tmp incremented.
tmp incremented.
tmp incremented.
tmp incremented.
tmp incremented.
Node word: [[word!]]
Node word: [[humongously-and-tempestuously-but-neither-abstemiously-nor-facetiously-long]]
Node word: [[stuvwxyz]]
Node word: [[pqr]]
Node word: [[mno]]
Node word: [[ghi]]
Node word: [[def]]
Node word: [[abc]]
Freeing list
Freeing: [[word!]]
Freeing: [[humongously-and-tempestuously-but-neither-abstemiously-nor-facetiously-long]]
Freeing: [[stuvwxyz]]
Freeing: [[pqr]]
Freeing: [[mno]]
Freeing: [[ghi]]
Freeing: [[def]]
Freeing: [[abc]]

当在Valgrind下运行时,我得到了一些奇怪的输出,但那是因为我已经升级了o / s(Mac OS X Yosemite到El Capitan)而没有更新抑制。泄漏都在系统代码中,而不是在此代码中,AFAICT。

后聊天

代码的一个特征是,如果输入两次单词,则该单词应移动到列表的前面。我的测试是关于独特的单词集。问题在于处理第一个单词的副本。这段代码中的修复程序看起来很健壮:

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>


typedef struct Node2
{
    char* word;
    struct Node2 *next;
} Node2;

Node2* head = NULL;
Node2* now = NULL;
int uniqueWords = 0;
int totalMTFRComparisons = 0;

bool DEBUG = true;

static
Node2* iNode(char* word)
{   
    if (DEBUG)
        printf("addNode2 called [[%s]].\n", word);
    Node2* tmp = head;
    Node2* prev = 0;

    while (tmp)
    {
        totalMTFRComparisons++;
        printf("Current word: %s\n", tmp->word);
        if (strcmp(tmp->word, word) == 0)
        {
            printf("Already entered: [[%s]]\n", tmp->word);
            if (prev != 0)
            {
                prev->next = tmp->next;
                tmp->next = head;
                head = tmp;
            }
            //tmp->next = prev;
            return tmp;
        }
        prev = tmp;
        tmp = tmp->next;
        printf("tmp incremented.\n");
    }

    Node2* ptr = malloc(sizeof(Node2));
    printf("New word: [[%s]]\n", word);
    uniqueWords++;
    ptr->word = malloc(strlen(word) + 1);
    strcpy(ptr->word, word);
    ptr->next = NULL;
    if (head == NULL)
    {
        head = now = ptr;
        return ptr;
    }
    else
    {
        ptr->next = head;
        head = ptr;
    }
    return ptr;
}

static
char* getString(void)
{
    static char buffer[128];
    while (fgets(buffer, 128, stdin) != NULL)
    {
        char *nl = strchr(buffer, '\n');
        if (nl != 0)
            *nl = '\0';
        printf("Line: [[%s]]\n", buffer);
        iNode(buffer);
    }
    return buffer;
}

int main(void)
{
    getString();
    printf("Total words: %d, total MTFR comparisons: %d\n", uniqueWords, totalMTFRComparisons);

    Node2* ptr2 = head;
    Node2* tmp2 = head;
    while (ptr2 != NULL)
    {
        printf("Freeing: [[%s]]\n", ptr2->word);
        tmp2 = ptr2->next;
        free(ptr2->word);
        free(ptr2);
        ptr2 = tmp2;
    }
}

它有一些诊断打印。新行剥离意味着诊断输出比您在聊天中看到的要短,如果您懒得查看聊天(字符串末尾有换行符 - 行被视为单词,包括换行符)。 / p>

答案 1 :(得分:0)

ptr->word = malloc(strlen(word));

以上行:不正确,记忆力很小Ex:strlen(&#34;你好&#34;)= 5而sizeof(&#34;你好&#34;)会给你6

将行更改为

ptr->word = malloc(sizeof(word));

然后strcpy将正常工作