Cats萌萌
这里没有什么复杂的,只要记住最后一个命令应该输出到原始进程'文件描述符1,第一个应该从原始进程文件描述符0读取。你只是按顺序生成进程,沿着输入端传输previois pipe电话。所以,以下是类型:#include <unistd.h>struct command{
const char **argv;};使用简单明确定义的语义创建一个辅助函数:intspawn_proc (int in, int out, struct command *cmd){
pid_t pid;
if ((pid = fork ()) == 0)
{
if (in != 0)
{
dup2 (in, 0);
close (in);
}
if (out != 1)
{
dup2 (out, 1);
close (out);
}
return execvp (cmd->argv [0], (char * const *)cmd->argv);
}
return pid;}这是主叉例程:intfork_pipes (int n, struct command *cmd){
int i;
pid_t pid;
int in, fd [2];
/* The first process should get its input from the original file descriptor 0. */
in = 0;
/* Note the loop bound, we spawn here all, but the last stage of the pipeline. */
for (i = 0; i < n - 1; ++i)
{
pipe (fd);
/* f [1] is the write end of the pipe, we carry `in` from the prev iteration. */
spawn_proc (in, fd [1], cmd + i);
/* No need for the write end of the pipe, the child will write here. */
close (fd [1]);
/* Keep the read end of the pipe, the next child will read from there. */
in = fd [0];
}
/* Last stage of the pipeline - set stdin be the read end of the previous pipe
and output to the original file descriptor 1. */
if (in != 0)
dup2 (in, 0);
/* Execute the last stage with the current process. */
return execvp (cmd [i].argv [0], (char * const *)cmd [i].argv);}还有一个小测试:intmain (){
const char *ls[] = { "ls", "-l", 0 };
const char *awk[] = { "awk", "{print $1}", 0 };
const char *sort[] = { "sort", 0 };
const char *uniq[] = { "uniq", 0 };
struct command cmd [] = { {ls}, {awk}, {sort}, {uniq} };
return fork_pipes (4, cmd);}似乎工作。:)