在我因为没有看这里过多的 Valgrind 问题而崩溃之前:我看了。我花了很多时间寻找，但我发现的都是没有 malloc() 正确字节数的人。如果我不正确并且这是重复的，我很乐意接受您的引用。
在这个程序中，我使用树来创建拼写检查器。我的代码中有很多问题需要修复，但内存泄漏是我真正需要帮助解决的唯一问题。 (很多代码都是临时的，所以它会编译，以便我可以修复内存泄漏。)
问题是我相当确定我正在为我的节点分配正确的空间量，我认为 Valgrind 证实了这一点，因为我只有 2 个未释放的 block (在 365,371 个分配中)。
无论如何，我会发布完整的代码(以防有人需要完整的上下文)，但我认为相关部分是加载函数和清除函数，我分别在其中分配和释放内存。
/**
* 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
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(确实是 dictionary.c:111 行)被执行两次，因此分配的内存永远不会被释放。 (这是正确的吗？)现在，这让我认为真正的问题在于我的 clear 函数，即它写得不好并且没有清除我的 trie 的每个节点。
但是，我盯着代码看了好几个小时，我几乎看不出它有什么问题。 (我敢肯定很多是；我只是不太擅长这个。)
如有任何帮助，我们将不胜感激。
最佳答案
对于初学者:
代码未将成员数组 children 初始化为每个节点 malloc() 的所有 NULL。 malloc() 不对分配的内存执行任何初始化。它只是包含“垃圾”。
所以这里
if (current_node->children[index] == NULL)
{
成为一个“随机”决定，如果它本身没有引发未定义的行为的话。
(顺便说一句，您的问题标题中提到的“无效读/写”，但您没有向我们展示，很可能也恰好提到了上面的这一行代码...)
要解决此问题，您可以使用如下内容:
#include <stdlib.h> (/* for malloc() */
#include <errno.h> (/* for errno */
/* Allocate and initialises a new node. */
/* Returns 0 on success and -1 on failure. */
int node_create(node ** pn)
{
int result = 0; /* Be optimistic. */
if (NULL == pn)
{
errno = EINVAL;
result = -1;
}
else
{
node * n = malloc(sizeof *n);
if (NULL == n)
{
result = -1;
}
else
{
for (size_t i = 0; i < LETTERS; ++i)
{
n -> children[i] = NULL;
}
n -> is_word = 0;
(*pn) = n;
}
}
return result;
}
然后像这样使用它:
...
if (-1 == node_create(&root))
{
perror("node_create() failed");
return false;
}
关于c - 内存泄漏和无效读/写，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39944782/