Valgrind错误：使用由struct和malloc构成的列表时，大小为8的无效写入

该函数cambiaDirezioneSinistra具有未定义的行为。

对于初学者，指向头节点的指针按值传递。因此，该函数处理指向头节点的指针的值的副本。更改副本不会影响原始指针。

void cambiaDirezioneSinistra(nodo_t * head)

指向头节点的指针可以等于NULL。但是函数内没有检查是否head等于NULL。

在此循环之后

for(tmp = head; tmp != NULL; tmp = tmp->next)
    ;

指针tmp将等于NULL。因此，使用此指针访问结构的数据成员没有任何意义。

可以通过以下方式声明和定义函数

//function with error
void cambiaDirezioneSinistra( nodo_t **head )
{
    if ( *head != NULL && ( *head )->next != NULL )
    {
        nodo_t *last = *head;

        while ( last->next ) last = last->next;

        last->next = *head;
        *head = ( *head )->next;

        last->next->next = NULL;
    }
}

这个函数叫做

cambiaDirezioneSinistra( &head );

这是一个演示程序。

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

//list' typedef
typedef struct nodo_s 
{
    int n;
    struct nodo_s * next;
} nodo_t;


//function with error
void cambiaDirezioneSinistra( nodo_t **head )
{
    if ( *head != NULL && ( *head )->next != NULL )
    {
        nodo_t *last = *head;

        while ( last->next ) last = last->next;

        last->next = *head;
        *head = ( *head )->next;

        last->next->next = NULL;
    }
}

int append( nodo_t **head, int n )
{
    while ( *head ) head = &( *head )->next;
    
    *head = malloc( sizeof( nodo_t ) );
    int success = *head != NULL;
    
    if ( success )
    {
        ( *head )->n = n;
        ( *head )->next = NULL;
    }
    
    return success;
}

void display( const nodo_t *head )
{
    for ( ; head != NULL; head = head->next )
    {
        printf( "%d -> ", head->n );
    }
    
    puts( "null" );
}

void clear( nodo_t **head )
{
    while ( *head )
    {
        nodo_t *tmp = *head;
        *head = ( *head )->next;
        free( tmp );
    }
}

int main(void) 
{
    nodo_t *head = NULL;
    
    const int N = 10;
    
    for ( int i = 0; i < N; i++ ) append( &head, i );
    
    display( head );
    
    for ( int i = 0; i < N; i++ )
    {
        cambiaDirezioneSinistra( &head );
    
        display( head );
    }
    
    clear( &head );
    
    return 0;
}

它的输出是

0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> null
2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> null
3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> null
4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> null
5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> null
6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null
7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null
8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null
9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

请注意，如果您需要访问函数中列表的最后一个节点，则最好将单链接列表定义为两侧的单链接列表。

例如

typedef struct lista_t
{
    nodo_t *head;
    nodo_t *tail;
} lista_t;