无法修复的内存泄漏

Question

对于很长的代码段，我深表歉意，但是我花了很多时间在这里看，我觉得到目前为止我所看到的任何东西都无法帮助我解决这个问题。 我在课程论坛上提出了问题，得到了助教的帮助，并从朋友那里得到了建议，但没有什么能够在这里找到问题的根源。

在此程序中，我使用树来创建拼写检查器。 我的代码中有很多事情需要修复，但是内存泄漏是我真正需要帮助解决的唯一问题。

问题是，我可以确定我为节点分配了正确的空间量，并且我认为Valgrind会确认这一点，因为我只有2个未释放的块（在365,371个分配中）。

无论如何，我将发布整个代码（以防万一任何人需要完整的上下文），但是我假定的相关部分分别是load函数和clear函数，分别在其中分配和释放内存。

/**
c* Implements a dictionary's functionality.
*/
#include <ctype.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#include "dictionary.h"

// number of characters we are using (a-z and ')
#define LETTERS 27

// max guaranteed number of nonnegative char values that exist
#define CHARVALUES 128

// create node structure for trie
typedef struct node
{
    struct node *children[LETTERS];
    bool is_word;
}
node;

// create root node for trie
node *root;

// stores the size of our dictionary
unsigned int dict_size = 0;

/**
 * Returns true if word is in dictionary else false.
 */
bool check(const char *word)
{
    // keeps track of where we are; starts with root for each new word
    node *current_node = root;

    while (*word != '\0')
    {

        // indices: 'a' -> 0, ..., 'z' -> 25, '\' -> 26
        int index = (tolower(*word) - 'a') % CHARVALUES;
        if (index >= LETTERS - 1)
        {
            // by assumption, the char must be '\'' if not '\n' or a letter
            index = LETTERS - 1;
        }

        // if the node we need to go to is NULL, the word is not here
        if (current_node->children[index] == NULL)
        {
            return false;
        }

        // go to the next logical node, and look at the next letter of the word
        current_node = current_node->children[index];
        word++;
    }
    return current_node->is_word;
}

/**
 * Loads dictionary into memory. Returns true if successful else false.
 */
bool load(const char *dictionary)
{

    FILE *inptr = fopen(dictionary, "r");
    if (inptr == NULL)
    {
        return false;
    }

    // allocate memory for the root node
    root = malloc(sizeof(node));

    // store first letter (by assumption, it must be a lowercase letter)
    char letter = fgetc(inptr);

    // stores indices corresponding to letters
    int index = 0;

    /**
     * we can assume that there is at least one word; we will execute the loop
     * and assign letter a new value at the end. at the end of each loop, due
     * to the inside loop, letter will be a newline; we know the EOF in the
     * dictionary follows a newline, so the loop will terminate appropriately
     */
    do
    {
        // keeps track of where we are; starts with root for each new word
        node *current_node = root; 

        // this loop will only execute if our character is a letter or '\''
        while (letter != '\n')
        {
            // indices: 'a' -> 0, ..., 'z' -> 25, '\' -> 26
            index = (letter - 'a') % CHARVALUES;
            if (index >= LETTERS - 1)
            {
                // by assumption, the char must be '\'' if not '\n' or a letter
                index = LETTERS - 1;
            }

            // allocate memory for a node if we have not done so already
            if (current_node->children[index] == NULL)
            {
                current_node->children[index] = malloc(sizeof(node));

                // if we cannot allocate the memory, unload and return false
                if (current_node->children[index] == NULL)
                {
                    unload();
                    return false;
                }

            }

            // go to the appropriate node for the next letter in our word
            current_node = current_node->children[index];

            // get the next letter
            letter = fgetc(inptr);
        }

        // after each linefeed, our current node represents a dictionary word
        current_node->is_word = true;
        dict_size++;

        // get the next letter
        letter = fgetc(inptr);
    }
    while (letter != EOF);

    fclose(inptr);

    // if we haven't returned false yet, then loading the trie must have worked
    return true;
}

/**
 * Returns number of words in dictionary if loaded else 0 if not yet loaded.
 */
unsigned int size(void)
{
    return dict_size;
}

void clear(node *head)
{
    for (int i = 0; i < LETTERS; i++)
    {
        if (head->children[i] != NULL)
        {
            clear(head->children[i]);
        }
    }
    free(head);
}

    /**
     * Unloads dictionary from memory. Returns true if successful else false.
     */
    bool unload(void)
    {
        clear(root);
        return true;
    }

相关的valgrind输出如下：

==18981== HEAP SUMMARY:
==18981==     in use at exit: 448 bytes in 2 blocks
==18981==   total heap usage: 365,371 allocs, 365,369 frees, 81,843,792 bytes allocated
==18981== 
==18981== 448 (224 direct, 224 indirect) bytes in 1 blocks are definitely lost in loss record 2 of 2
==18981==    at 0x4C2AB80: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==18981==    by 0x4011B0: load (dictionary.c:111)
==18981==    by 0x4008CD: main (speller.c:40)
==18981== 
==18981== LEAK SUMMARY:
==18981==    definitely lost: 224 bytes in 1 blocks
==18981==    indirectly lost: 224 bytes in 1 blocks
==18981==      possibly lost: 0 bytes in 0 blocks
==18981==    still reachable: 0 bytes in 0 blocks
==18981==         suppressed: 0 bytes in 0 blocks
==18981== 1 errors in context 3 of 11:
==18981== 
==18981== 
==18981== Invalid read of size 8
==18981==    at 0x40120C: load (dictionary.c:123)
==18981==    by 0x4008CD: main (speller.c:41)
==18981==  Address 0xb3fde70 is 16 bytes before a block of size 224 alloc'd
==18981==    at 0x4C2AB80: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==18981==    by 0x4011CB: load (dictionary.c:111)
==18981==    by 0x4008CD: main (speller.c:41)
==18981== 
==18981== 
==18981== 1 errors in context 4 of 11:
==18981== Invalid read of size 8
==18981==    at 0x4011E0: load (dictionary.c:114)
==18981==    by 0x4008CD: main (speller.c:41)
==18981==  Address 0xb3fde70 is 16 bytes before a block of size 224 alloc'd
==18981==    at 0x4C2AB80: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==18981==    by 0x4011CB: load (dictionary.c:111)
==18981==    by 0x4008CD: main (speller.c:41)
==18981== 
==18981== 
==18981== 1 errors in context 5 of 11:
==18981== Invalid write of size 8
==18981==    at 0x4011D4: load (dictionary.c:111)
==18981==    by 0x4008CD: main (speller.c:41)
==18981==  Address 0xb3fde70 is 16 bytes before a block of size 224 alloc'd
==18981==    at 0x4C2AB80: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==18981==    by 0x4011CB: load (dictionary.c:111)
==18981==    by 0x4008CD: main (speller.c:41)
==18981== 
==18981== 
==18981== 1 errors in context 6 of 11:
==18981== Invalid read of size 8
==18981==    at 0x4011B2: load (dictionary.c:109)
==18981==    by 0x4008CD: main (speller.c:41)
==18981==  Address 0xb3fde70 is 16 bytes before a block of size 224 alloc'd
==18981==    at 0x4C2AB80: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==18981==    by 0x4011CB: load (dictionary.c:111)
==18981==    by 0x4008CD: main (speller.c:41)

因此，我对此输出的解释是在以下代码块中：

        if (current_node->children[index] == NULL)
        {
            current_node->children[index] = malloc(sizeof(node));

            // if we cannot allocate the memory, unload and return false
            if (current_node->children[index] == NULL)
            {
                unload();
                return false;
            }

        }

malloc语句（实际上是行dictionary.c：111）执行两次，以使分配的内存永不释放。 （这是正确的吗？）现在，这使我认为真正的问题出在我的清晰函数上，即，它写得不好并且不能清除我的特里的每个节点。

但是，我已经盯着代码看了几个小时，从字面上看我看不出任何错误。 （我敢肯定很多；我对此不太擅长。）

任何帮助，将不胜感激。

附带说明：我有多个人（不是课程人员）告诉我应该将子级数组中的所有指针初始化为NULL，但是课程人员直接告诉我这是可选的，并且我已经对其进行了测试方式具有相同的结果。 我知道即使它在技术上可以“工作”，也可能是可移植的，但是我知道那不是我想要的解决方案，因为我知道还有其他根本原因（即导致它无法正常工作的一个根本原因）任何设备...）

再说一次，如果您能以任何方式帮助我解决这里的逻辑问题，我将不胜感激。 我一直试图弄清楚这一点，但无济于事。

Answer 1

root = malloc(sizeof(node));

这给出了一块未初始化的内存。

if (current_node->children[index] == NULL)

在这里，您假设内存已经初始化，而实际上是垃圾。

您需要在使用root之前初始化root的内容，或者使用calloc将其全部设置为零。

Answer 2

在将两个malloc（）语句都用calloc（）切换后（如其他人所建议的那样；这消除了您本应避免的许多valgrind错误），添加了一个小样本字典和以下简约main（）：

int main() {

    load("dict.txt");

    printf("Checked: %i\n", check("hello"));
    printf("Checked: %i\n", check("sdfsdf"));

    unload();

    return 0;
}

...您的代码运行得更加干净，并且没有任何内存泄漏：

==636== HEAP SUMMARY:
==636==     in use at exit: 0 bytes in 0 blocks
==636==   total heap usage: 15 allocs, 15 frees, 42,688 bytes allocated
==636==
==636== All heap blocks were freed -- no leaks are possible
==636==
==636== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 12 from 8)

但是，您将遇到的一个明显的泄漏是，如果您从load（）返回false，则不会释放文件指针。

编辑：当您将大写单词引入字典时，Valgrind（再次）开始引发各种错误。 因此，请集中精力进行调试。