c語言記憶體泄漏的原因是什么？

我在 c 中創建了一個鏈表，我遇到了這個記憶體泄漏。Valgrind 說問題的根源在第 15 行，但我仍然看不到問題所在。我覺得很愚蠢，不知道問題出在哪里，但如果你能幫助我，那就太好了！提前致謝！

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

typedef struct node
{
    int number;
    struct node *next;
}
node;


int main(void)
{
    node *list = malloc(sizeof(node));
    if (list == NULL)
    {
        return 1;
    }
    list = NULL;
    // create link list
    for (int i = 0; i < 3; i  )
    {
        node *n = malloc(sizeof(node));
        if(n == NULL)
        {
            free(list);
            return 1;
        }
        n->number = i 1;
        n->next = NULL;

        n->next = list;
        list = n;
    }
    //print link list
    for (node *tmp = list; tmp != NULL; tmp = tmp->next)
    {
        printf("%d - ", tmp->number);
    }
    printf("\n");
    //free link list
    while (list != NULL)
    {
        node *tmp = list->next;
        free(list);
        list = tmp;
    }


}

這是我的 valgrind 輸出：

==3895== Memcheck, a memory error detector
==3895== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==3895== Using Valgrind-3.15.0 and LibVEX; rerun with -h for copyright info
==3895== Command: ./temp
==3895== 
3 - 2 - 1 - 
==3895== 
==3895== HEAP SUMMARY:
==3895==     in use at exit: 16 bytes in 1 blocks
==3895==   total heap usage: 4 allocs, 3 frees, 64 bytes allocated
==3895== 
==3895== 16 bytes in 1 blocks are definitely lost in loss record 1 of 1
==3895==    at 0x483B7F3: malloc (in /usr/lib/x86_64-linux-gnu/valgrind/vgpreload_memcheck-amd64-linux.so)
==3895==    by 0x401168: main (temp.c:15)
==3895== 
==3895== LEAK SUMMARY:
==3895==    definitely lost: 16 bytes in 1 blocks
==3895==    indirectly lost: 0 bytes in 0 blocks
==3895==      possibly lost: 0 bytes in 0 blocks
==3895==    still reachable: 0 bytes in 0 blocks
==3895==         suppressed: 0 bytes in 0 blocks
==3895== 
==3895== For lists of detected and suppressed errors, rerun with: -s
==3895== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 0 from 0)

uj5u.com熱心網友回復：

指標只是一個存盤記憶體地址的變數。您使用 malloc 獲得了記憶體地址，malloc(sizeof(node))但您在其中輸入了 nulllist = NULL并且您丟失了地址。這不是你想要的嗎？

    node** list;
    size_t i;

    list = malloc(sizeof(node*));
    *list = NULL;

    for (i = 0; i < 3; i  )
    {
        node* n = malloc(sizeof(node));

        n->number = i   1;
        n->next = *list;

        *list = n;
    }

    /* print node... */
    /* release node... free(n) */
    free(list);
    list = NULL;

uj5u.com熱心網友回復：

很明顯，這種記憶體分配

node *list = malloc(sizeof(node));
if (list == NULL)
{
    return 1;
}
list = NULL;

是多余的，并且會產生記憶體泄漏，因為在list記憶體分配后重新分配了指標。結果，分配的記憶體地址丟失。

寫吧

node *list = NULL;

同樣在此代碼段中

    n->number = i 1;
    n->next = NULL;

    n->next = list;
    list = n;

這個說法

    n->next = NULL;

是多余的。

在 for 回圈中

for (int i = 0; i < 3; i  )
{
    node *n = malloc(sizeof(node));
    if(n == NULL)
    {
        free(list);
        return 1;
    }
    //...

您需要釋放所有分配的記憶體。

例如

for (int i = 0; i < 3; i  )
{
    node *n = malloc(sizeof(node));
    if(n == NULL)
    {
        while ( list )
        {
            node *tmp = list;
            list = list->next;
            free( tmp );
        } 
        return 1;
    }
    //...

為避免代碼重復，您可以撰寫一個單獨的函式來釋放分配的記憶體。例如

void free_list( node **list )
{
    while ( *list )
    {
        node *tmp = *list;
        *list = ( *list )->next; 
        free( tmp );
    }
}

并稱之為例如

for (int i = 0; i < 3; i  )
{
    node *n = malloc(sizeof(node));
    if(n == NULL)
    {
        free_list( &list );
        return 1;
    }
    //...

或者在程式結束時

free_list( &list );

標籤：C 结构 内存泄漏 动态内存分配 单链表

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