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某公司C语言笔试题,该如何解决

发布时间: 2012-04-06 12:22:24 作者: rapoo

某公司C语言笔试题
用C或者C++写一套通用函数。要求实现以下功能。
在一段连续内存(不大于10MB)中存储数据,要求FIFO。
至少有两个函数:
存入数据,将存入的数据放在缓冲区数据的尾部。第一参数为将要存入的数据首地址,第二参数为数据宽度。
取出数据,取出缓冲区最后的数据。第一参数为将要存入数据的指针,第二参数为数据宽度。
尽量提高效率。


[解决办法]
队列

queue.h

#ifndef _mytest_queue
#define _mytest_queue

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

typedef struct _mytest_QueueData
{
int num; // 存储在节点中的数据长度
char stu[100];
}mytest_QueueData;

typedef struct _mytest_Node
{
mytest_QueueData *data;
struct _mytest_Node *next; // 队列中的下一个节点地址
}mytest_NODE;

typedef struct _mytest_Queue
{
mytest_NODE *head; // 队列的头部
mytest_NODE *end; // 队列的尾部
int count; // 队列长度
}mytest_QUEUE;

void mytest_Init(mytest_QUEUE *queue )
{
if( NULL == queue )
return ;
queue->head = NULL;
queue->end = NULL;
queue->count = 0;
}

void mytest_Push(mytest_QUEUE *queue, mytest_QueueData *queue_data)
{
if( NULL == queue || NULL == queue_data )
return;
mytest_NODE *new_node = (mytest_NODE *)malloc(sizeof(mytest_NODE) );
if( NULL == new_node )
return;
new_node->data=(mytest_QueueData *)malloc(sizeof(mytest_QueueData));
if( NULL == new_node->data )
return;
memcpy(new_node->data,queue_data,sizeof(mytest_QueueData));
new_node->next = NULL;
if( queue->head == NULL )
{
queue->head = new_node;
queue->end = new_node;
}
else
{
queue->end->next = new_node;
queue->end = new_node;
}
queue->count ++;
}

mytest_QueueData *mytest_Front(mytest_QUEUE *queue )
{
if( NULL == queue )
return NULL;
if( NULL == queue->head )
return NULL;
mytest_QueueData *queuedata_tmp = queue->head->data;
return queuedata_tmp;
}

void mytest_Pop( mytest_QUEUE *queue )
{
if( NULL == queue )
return;
if( NULL == queue->head )
return;
mytest_NODE *node_tmp = queue->head;
mytest_QueueData *queuedata_tmp=node_tmp->data;
queue->head = node_tmp->next;
if( NULL == queue->head )
queue->end = NULL;
queue->count --;
free(queuedata_tmp);
free(node_tmp);
queuedata_tmp=NULL;
node_tmp=NULL;
}

void mytest_FreeQueue(mytest_QUEUE *queue )
{
if( queue )
return;
mytest_NODE *tmp_node1 = queue->head;
while(tmp_node1)
{
mytest_NODE *tmp_node2 = tmp_node1;
free(tmp_node1->data);
free(tmp_node1);
tmp_node1->data = NULL;
tmp_node1 = NULL;
tmp_node1 = tmp_node2->next;
}
}

#endif







test.c



#include <stdio.h>
#include "include/queue.h"
mytest_QUEUE qe;

void test()
{
mytest_QueueData qd;
int i;

for(i=0;i<1000000;i++)
{
sprintf(qd.stu,"%d:fffzcvz",i);
qd.num=i;
mytest_Push(&qe,&qd);
}
while(qe.count>0)
{
printf("%s\n",mytest_Front(&qe)->stu);
mytest_Pop(&qe);
}
printf("end1 \n");


}

int main()
{

mytest_Init(&qe);
int n=0;
while(n++<1000)
test();

printf("fff");
return 0;
}


[解决办法]
题目要求是在连续内存上操作,用链表不能满足要求吧。
[解决办法]
so easy.

记录一个尾指针,复制出去复制进来都用memcpy,整体代码不会超过50行的。
[解决办法]

[解决办法]
很明显是个字节流环形队列, 需要注意队列满的判定方法: 剩余内存小于 < 阈值, 则认为队列满.
[解决办法]
要求FIFO 是不是写错了.按照后面的加入,取出数据完全是栈。
[解决办法]
至少有两个函数:
存入数据,将存入的数据放在缓冲区数据的尾部。
取出数据,取出缓冲区最后的数据.

[解决办法]

探讨

引用:

so easy.

记录一个尾指针,复制出去复制进来都用memcpy,整体代码不会超过50行的。

我也做的是直接malloc 10MB内存,结果面试官说效率太低了。

[解决办法]
仅供参考
C/C++ code
//循环向a函数每次发送200个字节长度(这个是固定的)的buffer,//a函数中需要将循环传进来的buffer,组成240字节(也是固定的)的新buffer进行处理,//在处理的时候每次从新buffer中取两个字节打印#include <stdio.h>#include <stdlib.h>#include <string.h>#include <windows.h>#include <process.h>#include <io.h>//Log{#define MAXLOGSIZE 10000000#define ARRSIZE(x) (sizeof(x)/sizeof(x[0]))#include <time.h>#include <sys/timeb.h>#include <stdarg.h>char logfilename1[]="MyLog1.log";char logfilename2[]="MyLog2.log";char logstr[16000];char datestr[16];char timestr[16];char mss[4];CRITICAL_SECTION cs_log;FILE *flog;void Lock(CRITICAL_SECTION *l) {    EnterCriticalSection(l);}void Unlock(CRITICAL_SECTION *l) {    LeaveCriticalSection(l);}void LogV(const char *pszFmt,va_list argp) {    struct tm *now;    struct timeb tb;    if (NULL==pszFmt||0==pszFmt[0]) return;    if (-1==_vsnprintf(logstr,ARRSIZE(logstr),pszFmt,argp)) logstr[ARRSIZE(logstr)-1]=0;    ftime(&tb);    now=localtime(&tb.time);    sprintf(datestr,"%04d-%02d-%02d",now->tm_year+1900,now->tm_mon+1,now->tm_mday);    sprintf(timestr,"%02d:%02d:%02d",now->tm_hour     ,now->tm_min  ,now->tm_sec );    sprintf(mss,"%03d",tb.millitm);    printf("%s %s.%s %s",datestr,timestr,mss,logstr);    flog=fopen(logfilename1,"a");    if (NULL!=flog) {        fprintf(flog,"%s %s.%s %s",datestr,timestr,mss,logstr);        if (ftell(flog)>MAXLOGSIZE) {            fclose(flog);            if (rename(logfilename1,logfilename2)) {                remove(logfilename2);                rename(logfilename1,logfilename2);            }            flog=fopen(logfilename1,"a");            if (NULL==flog) return;        }        fclose(flog);    }}void Log(const char *pszFmt,...) {    va_list argp;    Lock(&cs_log);    va_start(argp,pszFmt);    LogV(pszFmt,argp);    va_end(argp);    Unlock(&cs_log);}//Log}#define ASIZE    200#define BSIZE    240#define CSIZE      2char Abuf[ASIZE];char Cbuf[CSIZE];CRITICAL_SECTION cs_HEX ;CRITICAL_SECTION cs_BBB ;struct FIFO_BUFFER {    int  head;    int  tail;    int  size;    char data[BSIZE];} BBB;int No_Loop=0;void HexDump(int cn,char *buf,int len) {    int i,j,k;    char binstr[80];    Lock(&cs_HEX);    for (i=0;i<len;i++) {        if (0==(i%16)) {            sprintf(binstr,"%03d %04x -",cn,i);            sprintf(binstr,"%s %02x",binstr,(unsigned char)buf[i]);        } else if (15==(i%16)) {            sprintf(binstr,"%s %02x",binstr,(unsigned char)buf[i]);            sprintf(binstr,"%s  ",binstr);            for (j=i-15;j<=i;j++) {                sprintf(binstr,"%s%c",binstr,('!'<buf[j]&&buf[j]<='~')?buf[j]:'.');            }            Log("%s\n",binstr);        } else {            sprintf(binstr,"%s %02x",binstr,(unsigned char)buf[i]);        }    }    if (0!=(i%16)) {        k=16-(i%16);        for (j=0;j<k;j++) {            sprintf(binstr,"%s   ",binstr);        }        sprintf(binstr,"%s  ",binstr);        k=16-k;        for (j=i-k;j<i;j++) {            sprintf(binstr,"%s%c",binstr,('!'<buf[j]&&buf[j]<='~')?buf[j]:'.');        }        Log("%s\n",binstr);    }    Unlock(&cs_HEX);}int GetFromRBuf(int cn,CRITICAL_SECTION *cs,FIFO_BUFFER *fbuf,char *buf,int len) {    int lent,len1,len2;    lent=0;    Lock(cs);    if (fbuf->size>=len) {        lent=len;        if (fbuf->head+lent>BSIZE) {            len1=BSIZE-fbuf->head;            memcpy(buf     ,fbuf->data+fbuf->head,len1);            len2=lent-len1;            memcpy(buf+len1,fbuf->data           ,len2);            fbuf->head=len2;        } else {            memcpy(buf     ,fbuf->data+fbuf->head,lent);            fbuf->head+=lent;        }        fbuf->size-=lent;    }    Unlock(cs);    return lent;}void thdB(void *pcn) {    char        *recv_buf;    int          recv_nbytes;    int          cn;    int          wc;    int          pb;    cn=(int)pcn;    Log("%03d thdB              thread begin...\n",cn);    while (1) {        Sleep(10);        recv_buf=(char *)Cbuf;        recv_nbytes=CSIZE;        wc=0;        while (1) {            pb=GetFromRBuf(cn,&cs_BBB,&BBB,recv_buf,recv_nbytes);            if (pb) {                Log("%03d recv %d bytes\n",cn,pb);                HexDump(cn,recv_buf,pb);                Sleep(1);            } else {                Sleep(1000);            }            if (No_Loop) break;//            wc++;            if (wc>3600) Log("%03d %d==wc>3600!\n",cn,wc);        }        if (No_Loop) break;//    }}int PutToRBuf(int cn,CRITICAL_SECTION *cs,FIFO_BUFFER *fbuf,char *buf,int len) {    int lent,len1,len2;    Lock(cs);    lent=len;    if (fbuf->size+lent>BSIZE) {        lent=BSIZE-fbuf->size;    }    if (fbuf->tail+lent>BSIZE) {        len1=BSIZE-fbuf->tail;        memcpy(fbuf->data+fbuf->tail,buf     ,len1);        len2=lent-len1;        memcpy(fbuf->data           ,buf+len1,len2);        fbuf->tail=len2;    } else {        memcpy(fbuf->data+fbuf->tail,buf     ,lent);        fbuf->tail+=lent;    }    fbuf->size+=lent;    Unlock(cs);    return lent;}void thdA(void *pcn) {    char        *send_buf;    int          send_nbytes;    int          cn;    int          wc;    int           a;    int          pa;    cn=(int)pcn;    Log("%03d thdA              thread begin...\n",cn);    a=0;    while (1) {        Sleep(100);        memset(Abuf,a,ASIZE);        a=(a+1)%256;        if (16==a) {No_Loop=1;break;}//去掉这句可以让程序一直循环直到按Ctrl+C或Ctrl+Break或当前目录下存在文件No_Loop        send_buf=(char *)Abuf;        send_nbytes=ASIZE;        Log("%03d sending %d bytes\n",cn,send_nbytes);        HexDump(cn,send_buf,send_nbytes);        wc=0;        while (1) {            pa=PutToRBuf(cn,&cs_BBB,&BBB,send_buf,send_nbytes);            Log("%03d sent %d bytes\n",cn,pa);            HexDump(cn,send_buf,pa);            send_buf+=pa;            send_nbytes-=pa;            if (send_nbytes<=0) break;//            Sleep(1000);            if (No_Loop) break;//            wc++;            if (wc>3600) Log("%03d %d==wc>3600!\n",cn,wc);        }        if (No_Loop) break;//    }}int main() {    InitializeCriticalSection(&cs_log );    Log("Start===========================================================\n");    InitializeCriticalSection(&cs_HEX );    InitializeCriticalSection(&cs_BBB );    BBB.head=0;    BBB.tail=0;    BBB.size=0;    _beginthread((void(__cdecl *)(void *))thdA,0,(void *)1);    _beginthread((void(__cdecl *)(void *))thdB,0,(void *)2);    if (!access("No_Loop",0)) {        remove("No_Loop");        if (!access("No_Loop",0)) {            No_Loop=1;        }    }    while (1) {        Sleep(1000);        if (No_Loop) break;//        if (!access("No_Loop",0)) {            No_Loop=1;        }    }    Sleep(3000);    DeleteCriticalSection(&cs_BBB );    DeleteCriticalSection(&cs_HEX );    Log("End=============================================================\n");    DeleteCriticalSection(&cs_log );    return 0;} 


[解决办法]

探讨

引用:

so easy.

记录一个尾指针,复制出去复制进来都用memcpy,整体代码不会超过50行的。

我也做的是直接malloc 10MB内存,结果面试官说效率太低了。

[解决办法]
探讨

C/C++ code
char data[10485760];

这样就够了 用全局静态的数组,比动态分配快

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