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Lab1 Xv6 and Unix utilities

2024-11-28 来源:个人技术集锦

Lab1 Xv6 and Unix utilities

sleep

user/sleep.c

#include "kernel/types.h"
#include "kernel/stat.h"
#include "user/user.h"

int
main(int argc, char *argv[])
{
	if(argc == 2){
		sleep(atoi((const char*) argv[1])); // convert ascii to integer and use system call sleep
		exit(0);
	}
	else{
		fprintf(2, "Usage:sleep number...\n");  // put the error masssage to the file descriptor 2
		exit(1); 	// terminate the process, 
				    //status = 1: indicates failure, and reported to wait() 
	}
}

pingpong

user/pingpong.c

  • Write a program that uses UNIX system calls to ‘‘ping-pong’’ a byte between two processes over a pair of pipes, one for each direction.
  • The parent should send a byte to the child; the child should print “< pid >: received ping”, where < pid > is its process ID, write the byte on the pipe to the parent, and exit;
    the parent should read the byte from the child, print “< pid >: received pong”, and exit.
  • Your solution should be in the file user/pingpong.c.
#include "kernel/types.h"
#include "kernel/stat.h"
#include "user/user.h"

int
main(int argc, char *argv[])
{
	int p1[2]; // parent -> child
	int p2[2]; // child -> parent
    char buffer[] = {'x'};
    int length = sizeof(buffer);
	if(argc != 1){
		fprintf(2, "Usage: pingpong\n");
		exit(1);
	}
	pipe(p1);
	pipe(p2);
	if(fork() == 0){ // child
		close(p1[1]);
		close(p2[0]);
		if(read(p1[0], buffer, length) == 1){  // child receives the byte
			printf("%d: received ping\n", getpid());
			write(p2[1], buffer, length);     // child sends the byte to child
			close(p2[1]);
			exit(0);
		}
		close(p2[1]);
		exit(1);			
	}else{  // parent
		close(p1[0]);
		close(p2[1]);
		write(p1[1], buffer,  length); // parent sends a byte to the child
		close(p1[1]);
		wait((int *)0);
		if(read(p2[0], buffer, length) == 1){  // parent receives the byte
			printf("%d: received pong\n", getpid());
		}
		exit(0);
	}
}

primes

user/primes.c

Write a concurrent version of prime sieve using pipes. This idea is due to Doug McIlroy, inventor of Unix pipes.
The picture halfway down this page and the surrounding text explain how to do it. Your solution should be in the file user/primes.c.

#include "kernel/types.h"
#include "user/user.h"

void
close_pipe(int *p) {
  close(p[0]);
  close(p[1]);
}

void
primes() {
  int n, p, len;
  int fd[2];

  // read the frist number from previous prime 
  if ((len = read(0, &n, sizeof(int))) <= 0 || n <= 0) {
    close(1);
    exit(0);
  }
  
  // print first number to console
  printf("prime %d\n", n);
  
  pipe(fd);
  if (fork() == 0) { // child process
    close(0);
    dup(fd[0]);  // link the std i/o 0 to fd[0] 
    close_pipe(fd);
    primes();
  } else {      // parent process
    close(1);
    dup(fd[1]); // link the std i/o 1 to fd[1]
    close_pipe(fd); 
    while ((len = read(0, &p, sizeof(int))) > 0 && p > 0) {  // read the number from the prime
      if (p % n != 0) {
        write(1, &p, sizeof(int));  // filter the number
      }
    }
    close(1);
    wait((int *)0);   
  } 
  exit(0);
}

int
main(void) {
  int i;
  int fd[2];
  
  pipe(fd);
  if (fork() == 0) {  // child process
    close(0);
    dup(fd[0]);  // link the std i/o 0 to the fd[0]
    close_pipe(fd);
    primes();
  } else {
    close(1);
    dup(fd[1]);  // link the std i/o 1 to the fd[1]
    close_pipe(fd);
    for (i = 2; i <= 35; i++) {
      write(1, &i, sizeof(int));
    }
    close(1);
    wait((int *)0);
  }
  exit(0);
}

find

user/find.c

Write a simple version of the UNIX find program: find all the files in a directory tree with a specific name. Your
solution should be in the file user/find.c.

#include "kernel/types.h"
#include "kernel/stat.h"
#include "user/user.h"
#include "kernel/fs.h"

char*
fmtname(char *path)  // 将路径格式化为文件名
{
  static char buf[DIRSIZ+1];
  char *p;

  // Find first character after last slash. 从字符串末尾开始遍历,直到遇到'/'或者到字符串的头部为止
  for(p=path+strlen(path); p >= path && *p != '/'; p--) ;
  p++;

  // Return blank-padded name.
  if(strlen(p) >= DIRSIZ)  // 字符串太长了
    return p;
  memmove(buf, p, strlen(p)+1);
  return buf;
}

void
find(char *path, char *file)
{
  char buf[512], *p;
  int fd;
  struct dirent de;
  struct stat st;

  if((fd = open(path, 0)) < 0){  // 路径不存在
    fprintf(2, "find: cannot open %s\n", path);
    return;
  }

  if(fstat(fd, &st) < 0){   // 获取不了该路径的状态
    fprintf(2, "find: cannot stat %s\n", path);
    close(fd);
    return;
  }

  switch(st.type){
  case T_FILE:  // 如果这是一个文件
    	if(strcmp(fmtname(path), file) == 0){
			printf("%s\n", path);
	}
    break;

  case T_DIR: // 如果这是一个目录
    if(strlen(path) + 1 + DIRSIZ + 1 > sizeof buf){  // 路径太长,缓冲区装不下
      printf("find: path too long\n");
      break;
    }
    strcpy(buf, path); 
    p = buf+strlen(buf);
    *p++ = '/';  // 在路径末尾添加'/'
    while(read(fd, &de, sizeof(de)) == sizeof(de)){  // 从该路径读取出文件名或者目录名
      if(de.inum == 0 || de.inum == 1)
        continue;
      if (strcmp(de.name, ".") == 0 || strcmp(de.name, "..") == 0)
        continue;
      memmove(p, de.name, DIRSIZ);  // 在路径后面添加文件名or目录
      p[DIRSIZ] = 0;      // the end of string is '\0'
      find(buf, file);
    }
    break;

  default:
    break;
  }
  close(fd);
}


int
main(int argc, char *argv[])
{
	if(argc != 3){
		fprintf(2, "Usage: find <path> <file>\n");
		exit(1);
	}
	find(argv[1], argv[2]);
	exit(0);
}

xargs

user/xargs.c

Write a simple version of the UNIX xargs program: read lines from the standard input and run a command for each line, supplying the line as arguments to the command.
Your solution should be in the file user/xargs.c.

#include "kernel/types.h"
#include "user/user.h"
#include "kernel/param.h"

int
main(int argc, char *argv[])
{
	char *arguments[MAXARG];
	char buff[255];
	char *p = buff, *q = buff;
	int i, offset;
	if(argc <= 1){
		fprintf(2, "Usage: xargc <command>...\n");
		exit(1);
	}
	
	for(i = 1; i < argc; i++){  // get the command which is behind the "xargs"
        arguments[i-1] = argv[i];
    }
    i = i - 1;
    offset = i;  // backup
    
    while(read(0, p, 1)){
    	if(*p == '\n'){  // meet the '\n' and execute it with arguments
    		*p = '\0';
    		arguments[i] = q;
    		if(fork() == 0){
    			exec(arguments[0], arguments);
    			exit(0);
    		}else{
    			wait((int*)0);
    			p = buff;
    			q = p;
    			i = offset;
    		}
    	}else if(*p == ' ' || *p == '\t'){  // meet the ' ' or '\t', find an argument
    		*p = '\0';
    		arguments[i] = q;
    		i = i + 1;
    		p = p + 1;
    		q = p; 
    	}else{
    		p = p + 1;
    	}
    }
	exit(0);
}

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