操作系统：模拟内存回收算法

实例:某系统内存容量为800K，下 面分别给出中的空闲分区表和分配分区表，系统采用动
态分区存储管理策略。现有按照如下顺序释放作业空间:

(1)释放作业:Task2[情况D];

(2)释放作业:Task4 [情况A];

(3)释放作业:Task6[情况B];

(4)释放作业:Task7[情况C];

(5)释 放作业:Task9 [不存在]

并打印每次回收分配空间后的空闲分区表和分配分区表。

1.动态地任意顺序输入“分配分区表项”，按照地址递增方式建立“分配分区表”：

输入：

2. 动态地任意顺序输入“空闲分区表项"”，按照地址递增方式建立“空闲分区表”

输入：

 3.共内存回收5次：

输入：

 4.回收内存分配区，模拟第一次内存回收：TASK2 【情况D】

输入：

分配成功，输出：

 回收内存分配区，模拟第二次内存回收: Task4 , 【情况A】。

输入：

分配成功，输出：

 回收内存分配区，模拟第三次内存回收: Task6 [ 情况B]。

输入：

 分配失败，输出：

 回收内存分配区，模拟第四次内存回收: Task7 , 【情况C】。

输入：

分配成功，输出：

 回收内存分配区，模拟第五次内存回收: Task9[ 不存在 ]。

输入：

 分配成功，输出：

#include 
#include 
#include 
#define NULL 0
typedef struct table
{
    int address; 
    int length; 
    int flag; 
    char name[10]; 
    struct table *next;
} node;
node *work; 
char type; 
bool success; 
void insert(node *head, node *p) 
{
    node *q,*w;
    if(head->next->address>p->address)
    {
        p->next=head->next;
        head->next=p;
        return;
    }
    q=head->next->next;
    w=head->next;
    while(q)
    {
        if(q->address>p->address)
            break;
        w=q;
        q=q->next;
    }
    w->next=p;
    p->next=q;
    return;
}
node *creat() 
{
    printf("address length flag(0 or 1):n");
    node *head,*p;
    head=NULL;
    head=(node *)malloc(sizeof(node));
    head->next=NULL;
    int a,b,c;
    if(head->next==NULL)
    {
        p=(node *)malloc(sizeof(node));
        scanf("%d%d%d",&a,&b,&c);
        if(a==0&&b==0)
            return head;
        p->address=a;
        p->length=b;
        p->flag=c;
        if(p->flag==1)
        {
            printf("tinput job_name:");
            scanf("%s",p->name);
        }
        else
            strcpy(p->name,"nil");
        head->next=p;
        head->next->next=NULL;
    }
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        if(a==0&&b==0&&c==0)
            break;
        p=(node *)malloc(sizeof(node));
        p->address=a;
        p->length=b;
        p->flag=c;
        if(p->flag==1)
        {
            printf("tinput job_name:");
            scanf("%s",p->name);
        }
        else
            strcpy(p->name,"nil");
        insert(head,p);
    }
    return head;
}
node *found(node *distributedhead,char workn[10]) 
{
    success=false;
    node *p,*q,*w;
    p=distributedhead->next;
    q=distributedhead;
    w=(node *)malloc(sizeof(node));
    w=NULL;
    while(p)
    {
        if(strcmp(p->name,workn)==0)
            break;
        q=p;
        p=p->next;
    }
    if(p)
    {
        success=true;
        w=p;
        w->flag=0;
        strcpy(w->name,"nil");
        q->next=p->next;
    }
    return w;
}
void release(node *freehead,node *work) 
{
    if(freehead->next->address>(work->address+work->length))
    {
        printf("nThe type of release is D!n");
        work->next=freehead->next;
        freehead->next=work;
        return;
    }
    else if(freehead->next->address==(work->address+work->length))
    {
        printf("nThe type of release is A!n");
        freehead->next->address=work->address;
        freehead->next->length+=work->length;
        return;
    }
    node *p;
    p=freehead->next;
    while(p)
    {
        if(p->next&&work->address==(p->address+p->length)&&(work->address+work->length)next->address)
        {
            printf("nThe type of release is A!n");
            p->length+=work->length;
            return;
        }
        if(p->next&&work->address==(p->address+p->length)&&(work->address+work->length)==p->next->address)
        {
            printf("nThe type of release is C!n");
            p->length+=p->next->length+work->length;
            p->next=p->next->next;
            return;
        }
        if(p->next&&work->address>(p->address+p->length)&&(work->address+work->length)==p->next->address)
        {
            printf("nThe type of release is B!n");
            p->next->address-=work->length;
            p->next->length+=work->length;
            return;
        }
        if(p->next&&work->address>(p->address+p->length)&&(work->address+work->length)next->address)
        {
            printf("nThe type of release is D!n");
            work->next=p->next;
            p->next=work;
            return;
        }
        p=p->next;
    }
    work->next=(node *)malloc(sizeof(node));
    work->next=NULL;
    p=work;
    return;
}
void print (node *head) 
{
    node *p;
    p=head->next;
    while(p)
    {
        printf("%d,%d,%d,%sn",p->address,p->length,p->flag,p->name);
        p=p->next;
    }
}

int main()
{
    int i,sum;
    struct table *dtable,*ftable;
    char workn[10];
    printf("The distributed table is:n");
    dtable=creat(); 
    printf("The free table is:n");
    ftable=creat(); 
    
    printf("Input the released work segment sum:");
    scanf("%d",&sum);
    i=1;
    while(sum--)
    {
        printf("%d: input the released work segment name:",i++);
        scanf("%s",workn);
        work=(node *)malloc(sizeof(node));
        work=found(dtable,workn);
        if(success)
        {
            printf("n要回收的分区存在！n");
            release(ftable,work);
            printf("ndistributed table is:n");
            print(dtable);
            printf("nfree table is:n");
            print(ftable);
        }
        else
        {
            printf("n要回收的分区不存在！n");
        }
    }
    return 0;
}