为分配创建自己的 malloc。获得分段错误

Question

分段错误发生在有注释的点。我认为这与我没有初始化头部和尾部节点这一事实有关。我也尝试将其初始化为 NULL，但没有成功。不幸的是，我真的不知道如何在不使用 malloc 的情况下初始化它们。任何帮助都会很棒。谢谢。

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

//the structure of the node in the linked list
typedef struct Node{
    int size;
    int status;
    struct Node* next;
    struct Node* previous;
}Node;

int* HEAP_START = 0;
int* HEAP_END = 0;
Node* head;
Node* tail;

int first = 0;
//printf("here1\n");

void *my_bestfit_malloc(int size)
{
    Node* newNode = NULL;
    printf("here2\n");
    if(first == 0)
    {
        HEAP_START = (int*)sbrk(0);
        newNode = sbrk(size + sizeof(Node));
        HEAP_END = (int*)sbrk(0);
        head->next = tail;      //segmentation error happens here
        printf("here3\n");
        tail->previous = head;
        newNode->size = size;
        newNode->status = 1;
        first++;
    }
    else
    {
        Node* currNode = head->next;
        printf("here4\n");
        while(currNode->next != tail)
        {
            if(currNode->size == size)
            {
                newNode = currNode;
                currNode->previous->next = currNode->next;
                currNode->next->previous = currNode->previous;
                newNode->size = size;
                newNode->status = 1;
                printf("here5\n");
                break;
            }
            else
            {
                currNode = currNode->next;
                printf("here6\n");
            }
        }
        if(currNode->next == tail)
        {
            newNode = sbrk(size + sizeof(Node));
            HEAP_END = (int*)sbrk(0);
            newNode->size = size;
            newNode->status = 1;
            printf("here7\n");
        }
    }
    return newNode + sizeof(Node);
}

 int main()
{
    typedef struct person{
        int age;
        char sex;
    }person;
    printf("main1\n");
    person* dave = (person*)my_bestfit_malloc(sizeof(person));
    printf("main2\n");
    person* vicki = (person*)my_bestfit_malloc(sizeof(person));
    printf("main3");
    person* alex = (person*)my_bestfit_malloc(sizeof(person));

    dave->age = 26;
    dave->sex = 'M';

    vicki->age = 24;
    vicki->sex = 'F';

    alex->age = 19;
    alex->sex = 'F';

    printf("Dave:\n\tAge: %d\n\tSex: %c\n", dave->age, dave->sex);
    printf("Vicki:\n\tAge: %d\n\tSex: %c\n", dave->age, dave->sex);
    printf("Alex:\n\tAge: %d\n\tSex: %c\n", dave->age, dave->sex);
}

所以我尝试将我的 Node* 头部和尾部更改为：Node head;节点尾；相反，但收到以下错误：

mymalloc.c: In function ‘my_bestfit_malloc’:
mymalloc.c:38: error: invalid type argument of ‘->’ (have ‘Node’)
mymalloc.c:40: error: invalid type argument of ‘->’ (have ‘Node’)
mymalloc.c:47: error: invalid type argument of ‘->’ (have ‘Node’)
mymalloc.c:49: error: invalid operands to binary != (have ‘struct Node *’     and ‘Node’)
mymalloc.c:67: error: invalid operands to binary == (have ‘struct Node *’     and ‘Node’)

前三个我明白了，我需要用head.next = tail;相反，但我不明白最后两个。

最终编辑： 想通了就知道了。 head 和 tail 的指针需要是实际的 Node 结构而不是结构指针。我还需要返回一个 void 指针而不是一个节点。

Answer 1

看起来您正在为 head 赋值。 head 尚未分配。

改变：

Node* head;
Node* tail;

到：

Node head;
Node tail;

当访问指向struct use -> 的指针成员时。如head->size 而当访问一个struct use .的成员时如head.size。

我对您的代码进行了一些更改，现在程序至少可以分配和使用第一个person。但是，person 类型的后续分配失败。我怀疑void *my_bestfit_malloc(int size) 可能有一些逻辑/指针问题....

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

//the structure of the node in the linked list
typedef struct Node{
    int size;
    int status;
    struct Node* next;
    struct Node* previous;
}Node;

int* HEAP_START = 0;
int* HEAP_END = 0;
struct Node head;
struct Node tail;

int first = 0;
//printf("here1\n");

void *my_bestfit_malloc(int size)
{
    Node* newNode = NULL;
    printf("here2\n");
    if(first == 0)
    {
        HEAP_START = (int*)sbrk(0);
        newNode = sbrk(size + sizeof(Node));
        HEAP_END = (int*)sbrk(0);
        head.next = &tail;      //segmentation error happens here
        printf("here3\n");
        tail.previous = &head;
        newNode->size = size;
        newNode->status = 1;
        first++;
    }
    else
    {
        Node* currNode = head.next;
        printf("here4\n");
        while( currNode != &tail)
        {
            if(currNode->size == size)
            {
                newNode = currNode;
                currNode->previous->next = currNode->next;
                currNode->next->previous = currNode->previous;
                newNode->size = size;
                newNode->status = 1;
                printf("here5\n");
                break;
            }
            else
            {
                currNode = currNode->next;
                printf("here6\n");
            }
        }
        if(currNode->next == &tail)
        {
            newNode = sbrk(size + sizeof(Node));
            HEAP_END = (int*)sbrk(0);
            newNode->size = size;
            newNode->status = 1;
            printf("here7\n");
        }
    }
    return newNode + sizeof(Node);
}

 int main()
{
    typedef struct person{
        int age;
        char sex;
    }person;
    printf("main1\n");
    person* dave = (person*)my_bestfit_malloc(sizeof(person));
    printf("main2\n");
    person* vicki = (person*)my_bestfit_malloc(sizeof(person));
    printf("main3");
    person* alex = (person*)my_bestfit_malloc(sizeof(person));

    dave->age = 26;
    dave->sex = 'M';

    //vicki->age = 24;
    //vicki->sex = 'F';

    //alex->age = 19;
    //alex->sex = 'F';

    printf("Dave:\n\tAge: %d\n\tSex: %c\n", dave->age, dave->sex);
    //printf("Vicki:\n\tAge: %d\n\tSex: %c\n", dave->age, dave->sex);
    //printf("Alex:\n\tAge: %d\n\tSex: %c\n", dave->age, dave->sex);
}

现在运行代码至少会给出以下输出：

jrn@VirtualBox-mint17 ~ $ ./a.out 
main1
here2
here3
main2
here2
here4
main3here2
here4
Dave:
    Age: 26
    Sex: M