Linux课程大作业(共21页).doc

精选优质文档-倾情为你奉上Linux课程设计报告 题 目 Linux课程大作业院 系 班 级 姓 名 指导教师 一、基础篇（给出源程序和编译运行的结果）1、 编写一个简单的c语言程序：根据输入的两个整数求平均值并且在终端输出，通过gcc编译器得到它的汇编程序文件。源代码（c）：#include<stdio.h>double average( int m,int n)return (m+n)/2.0;int main(void)int m,n=0; printf("请输入两个数，回车分割n"); scanf("%d",&m); scanf("%d",&n); printf("%d与%d的平均值是：%lfn",m,n,average(m,n);源代码（汇编）：.file"sum.c".text.globlaverage.typeaverage, functionaverage:.LFB0:.cfi_startprocpushl%ebp.cfi_def_cfa_offset 8.cfi_offset 5, -8movl%esp, %ebp.cfi_def_cfa_register 5subl$8, %espmovl12(%ebp), %eaxmovl8(%ebp), %edxaddl%edx, %eaxmovl%eax, -4(%ebp)fildl-4(%ebp)fldl.LC0fdivrp%st, %st(1)leave.cfi_restore 5.cfi_def_cfa 4, 4ret.cfi_endproc.LFE0:.sizeaverage, .-average.section.rodata.align 4.LC2:.string"350257267350276223345205245344270244344270252346225260357274214345233236350275246345210206345211262".LC3:.string"%d".LC4:.string"%d344270216%d347232204345271263345235207345200274346230257357274232%lfn".text.globlmain.typemain, functionmain:.LFB1:.cfi_startprocpushl%ebp.cfi_def_cfa_offset 8.cfi_offset 5, -8movl%esp, %ebp.cfi_def_cfa_register 5andl$-16, %espsubl$48, %espmovl$0, 44(%esp)movl$.LC2, (%esp)callputsmovl$.LC3, %eaxleal40(%esp), %edxmovl%edx, 4(%esp)movl%eax, (%esp)call_isoc99_scanfmovl$.LC3, %eaxleal44(%esp), %edxmovl%edx, 4(%esp)movl%eax, (%esp)call_isoc99_scanfmovl44(%esp), %edxmovl40(%esp), %eaxmovl%edx, 4(%esp)movl%eax, (%esp)callaveragemovl44(%esp), %ecxmovl40(%esp), %edxmovl$.LC4, %eaxfstpl12(%esp)movl%ecx, 8(%esp)movl%edx, 4(%esp)movl%eax, (%esp)callprintfleave.cfi_restore 5.cfi_def_cfa 4, 4ret.cfi_endproc.LFE1:.sizemain, .-main.section.rodata.align 8.LC0:.long0.long.ident"GCC: (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3".section.note.GNU-stack,"",progbits执行结果：2、 编写一个c语言程序：打印输出所有“水仙花数”，用gdb调试程序（给出步骤，至少十步以上）。所谓“水仙花数”是指一个3位数，其各位数字立方和等于该数本身。例如，153是一水仙花数，因为153=1³+5³+3³。源代码：#include<stdio.h>int isnumber(int number) int ge=(number%100)%10; int shi=(number/10)%10; int bai=(number/100); printf("%dn",ge); printf("%dn",shi); printf("%dn",bai); int temp=ge*ge*ge+shi*shi*shi+bai*bai*bai; if(temp=number) return 1; elsereturn 0;int main()int number=0;printf("input number!n");scanf("%d",&number);if(number>1000&&number<0)printf("number limits outbounds!n");if(isnumber(number)printf("number is format!n");elseprintf("number is not format!n");return 0;执行结果：3、 设计一个程序，要求输出n个整数（n也由键盘输入）中的最大值，并为它编写makefile文件，用make编译后修成返回最小值，再编译，观察有多少文件不需要重新编译。源代码：max.cdouble max(double m,double n)if(m<n) return n; elsereturn m; main.c#include<stdio.h>#include"define.h"int main(void)double m,n=0; printf("请输入两个实数、n"); scanf("%lf",&m); scanf("%lf",&n); printf("%lf与%lf最大值是：%lfn",m,n,max(m,n);define.hdouble max(double m,double n);makefile文件：test:max.o main.ogcc max.o main.o -o testmain.o:main.c define.hgcc main.c -cmax.o:max.cgcc max.c -c结果图： 改变程序后，只有max.c，main.c文件程序需要重新编译；4、 编写一个程序，求2-n间的素数，n由键盘输入，循环变量分别从2到n、2到(int)sqrt(n)，分别测出两个循环的所用时间。源代码：#include "stdio.h"#include "math.h"int main()int t1=0,t2=0;int n;int j,k,j2,k2;int i,i2;printf("输入的值：");scanf("%d",&n);for(i=2;i<=n;i+)t1+;k=sqrt(i);for(j=2;j<k+1;j+)if(i%j=0)break;if(j>=k+1)printf("%d是素数n",i);printf("nn循环时间%d",t1);for(i2=2;i2<=(int)sqrt(n);i2+)t2+;k2=sqrt(i2);for(j2=2;j2<k2+1;j2+)if(i2%j2=0)break;if(j2>=k+1)printf("%d是素数n",i2);printf("nsqrt(n)循环时间%dn",t2);return 0;编译运行效果：5、设计一个程序，要求将10分别以十进制、八进制和十六进制输出。程序设计：#include<stdio.h>int main()int number=10;printf("十进制值：%dn",number);printf("八进制值：%on",number);printf("十六进制值：%xn",number); 运行结果：一、 提高篇（三选二，划出程序流程图，给出程序源代码和编译运行的结果）1、 设计两个程序，要求用命名管道FIFO，实现简单的文本文件或图片文件的传输功能。流程图： 源代码：send.c#include <stdio.h>#include <signal.h>#include <stdlib.h>#include <unistd.h>#include <sys/types.h>#include <sys/stat.h>#include <fcntl.h>#include <string.h>int main()char s128;int fd;FILE *fp;fp = fopen("./a.txt", "r");mkfifo("/tmp/fifo.tst", 0644);fd = open("/tmp/fifo.tst", O_WRONLY);while(fgets(s, 127, fp) != NULL) write(fd, s, strlen(s);printf("%s",s);close(fd);fclose(fp);unlink("/tmp/fifo.tst");return 0;get.c/*get.c*/#include <stdio.h>#include <signal.h>#include <stdlib.h>#include <unistd.h>#include <sys/types.h>#include <sys/stat.h>#include <string.h>#include <fcntl.h>int main()char s128;int fd = open("/tmp/fifo.tst", O_RDONLY);int fd2 = open("./b.txt", O_CREAT|O_WRONLY);memset(s, 0, 128); while(read(fd, s, 128) > 0) printf("%s", s); write(fd2, s, 128); printf("fd2=%dn",fd2);close(fd2);close(fd); return 0;运行结果：2、 设计两个程序，要求用消息队列，实现聊天程序，每次发言后自动在后面增加当前系统时间。增加结束字符，比如最后输入“88”后结束进程。流程图：发送源源代码：#include<stdio.h>#include<stdlib.h>#include<sys/ipc.h>#include<sys/msg.h>#include<string.h>struct msgbufint type;char ptr0;int main(int argc,char *argv)key_t key;key=ftok(argv1,100);int msgid;msgid=msgget(key,IPC_CREAT|0600);pid_t pid;pid=fork();if(pid=0)while(1)printf("pls input msg to send:");char buf128;fgets(buf,128,stdin);struct msgbuf *ptr=malloc(sizeof(struct msgbuf)+strlen(buf)+1);ptr->type=1;memcpy(ptr->ptr,buf,strlen(buf)+1);msgsnd(msgid,ptr,strlen(buf)+1,0);free(ptr);elsestruct msgbufint type;char ptr1024;while(1)struct msgbuf mybuf;memset(&mybuf,'0',sizeof(mybuf);msgrcv(msgid,&mybuf,1024,2,0);printf("recv msg:%sn",mybuf.ptr);接收端源代码：#include<stdio.h>#include<stdlib.h>#include<sys/ipc.h>#include<sys/msg.h>#include<string.h>struct msgbufint type;char ptr0;int main(int argc,char *argv)key_t key;key=ftok(argv1,100);int msgid;msgid=msgget(key,IPC_CREAT|0600);pid_t pid;pid=fork();if(pid=0)/sendwhile(1)printf("pls input msg to send:");char buf128;fgets(buf,128,stdin);struct msgbuf *ptr=malloc(sizeof(struct msgbuf)+strlen(buf)+1);ptr->type=2;/send msg type=2memcpy(ptr->ptr,buf,strlen(buf)+1);msgsnd(msgid,ptr,strlen(buf)+1,0);free(ptr);elsestruct msgbufint type;char ptr1024;while(1)struct msgbuf mybuf;memset(&mybuf,'0',sizeof(mybuf);msgrcv(msgid,&mybuf,1024,1,0);/recv msg type=2printf("recv msg:%sn",mybuf.ptr);运行结果：3、设计两个程序，要求用mmap系统，实现简单的聊天程序。二、 应用篇给出程序设计思路和程序设计流程图，并给出程序源代码和编译运行的结果。设计思路： 要求生产者-消费者在固定的仓库空间条件下，生产者每生产一个产品将占用一个仓库空间，生产者生产的产品库存不能越过仓库的存储量，消费者每消费一个产品将增加一个仓库空间，消费者在仓库产品为零时不能再消费。以下使用了两个信号量，一个用来管理消费者（sem_produce）,一个用来管理生产者（sem_custom）,sem_produce表示当前仓库可用空间的数量，sem_custom表示当前仓库中产品的数量。对于生产者来说，其需要申请的资源为仓库中的剩余空间，因此，生产者在生产一个产品前需申请sem_produce信号量，当此信号量的值大于零时，即有可用空间，将生产产品，并将sem_produce值减1，同时，当其生产一个产品后，当前仓库的产品数量加1，需要将sem_custom的值自动加1.对于消费者来说，其需要申请的资源为仓库中的产品，因此，消费者在消费一个产品前需申请sem_custom信号量，当此信号量的值大于零时，即有可用产品，将消费一个产品，并将sem_custom值减1，同时，当其消费一个产品后，当前仓库的剩余空间加1，需要sem_produce的值自动加。1、 生产者-消费者问题模拟程序。流程图：生产者源代码：#include <unistd.h>#include <stdlib.h>#include <stdio.h>#include <sys/types.h>#include <sys/ipc.h>#include <sys/sem.h>#include <errno.h>int sem_id;void init()key_t key;int ret;unsigned short sem_array2;union semun int val;struct semid_ds *buf;unsigned short *array;arg;key=ftok(".",'s');sem_id=semget(key,2,IPC_CREAT|0644);sem_array0=0;/identify the productorsem_array1=100;/identify the space/printf("set the productor init value is 0nset the space init value is 100n");arg.array = sem_array;ret = semctl(sem_id, 0, SETALL, arg);if (ret = -1) printf("SETALL failed (%d)n", errno); /printf("nread the numbern");printf("productor init is %dn",semctl(sem_id,0,GETVAL);printf("space init is %dnn",semctl(sem_id,1,GETVAL);void del()semctl(sem_id,IPC_RMID,0);int main(int argc,char *argv)struct sembuf sops2;sops0.sem_num = 0;sops0.sem_op = 1;sops0.sem_flg = 0;sops1.sem_num = 1;sops1.sem_op = -1;sops1.sem_flg = 0;init();printf("this is productorn");while(1)printf("nnbefore produce:n");printf("productor number is %dn",semctl(sem_id,0,GETVAL);printf("space number is %dn",semctl(sem_id,1,GETVAL);semop(sem_id,(struct sembuf *)&sops1,1);/get the space to instore the productorprintf("now producing.n");semop(sem_id,(struct sembuf *)&sops0,1);/now tell the customer can bu cusumeprintf("nafter producen");printf("spaces number is %dn",semctl(sem_id,1,GETVAL);printf("productor number is %dn",semctl(sem_id,0,GETVAL);sleep(4);del();消费者源代码：#include <unistd.h>#include <stdlib.h>#include <stdio.h>#include <sys/types.h>#include <sys/ipc.h>#include <sys/sem.h>#include <errno.h>int sem_id;void init()key_t key;key=ftok(".",'s');sem_id=semget(key,2,IPC_CREAT|0644);/printf("sem id is %dn",sem_id);int main(int argc,char *argv)init();struct sembuf sops2;sops0.sem_num = 0;sops0.sem_op = -1;sops0.sem_flg = 0;sops1.sem_num = 1;sops1.sem_op = 1;sops1.sem_flg = 0;init();printf("this is customern");while(1)printf("nnbefore consume:n");printf("productor is %dn",semctl(sem_id,0,GETVAL);printf("space is %dn",semctl(sem_id,1,GETVAL);semop(sem_id,(struct sembuf *)&sops0,1);/get the productor to cusumeprintf("now consuming.n");semop(sem_id,(struct sembuf *)&sops1,1);/now tell the productor can bu produceprintf("nafter consumen");printf("products number is %dn",semctl(sem_id,0,GETVAL);printf("space number is %dn",semctl(sem_id,1,GETVAL);sleep(3);编译运行结果：3、实现基于GTK+的网络聊天室，实现通信双方的实时通信。设计思想：初始化聊天窗口界面，建立socket连接，实现一对一的通信。流程图：专心-专注-专业