gpt4 book ai didi

c - (在C 中模拟UNIX SHELL)如何在for 循环中实现多个管道?

转载 作者:太空宇宙 更新时间:2023-11-03 23:37:30 25 4
gpt4 key购买 nike

我正在尝试在 C 程序中模拟 unix shell,它仍处于开始阶段,最多只能用于两个管道。我有一个命令 vector (char *com[3][3]),考虑到字符“|”,它们是分开的,但我的问题是如何在 for 循环中处理更多管道?按照当前的实现,我尝试执行 3 个由管道分隔的命令:

#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/wait.h>
#include <sys/types.h>

int main(int argc, char **argv){

//Vector with positions of pipes found, position 0 reserved for the total amount of commands.

char* com[3][3] = { { "/bin/ls", "-la", 0 },
{ "/bin/grep", ".", 0}, { "/usr/bin/wc", "-l", 0 }};

//EXECUTE COMMANDS
pid_t fork1, fork2, fork3;
int fd1[2], fd2[2];

if(pipe(fd1) < 0){
perror("pipe1");
}
if(pipe(fd2) < 0){
perror("pipe2");
}



//COMMAND 1
fork1 = fork();
if(fork1 == 0){
dup2(fd1[1], STDOUT_FILENO);
close(fd1[0]);
close(fd2[0]);
close(fd2[1]);
execvp(com[0][0], com[0]);
perror("execvp 1");
exit(EXIT_FAILURE);
}

//COMMAND 2
fork2 = fork();
if(fork2 == 0){
dup2(fd1[0], STDIN_FILENO);
dup2(fd2[1], STDOUT_FILENO);
close(fd1[1]);
close(fd2[0]);
execvp(com[1][0], com[1]);
perror("execvp 2");
exit(EXIT_FAILURE);
}

//COMMAND 3
fork3 = fork();
if(fork3 == 0){
dup2(fd2[0], STDIN_FILENO);
close(fd2[1]);
close(fd1[0]);
close(fd1[1]);
execvp(com[2][0], com[2]);
perror("execvp 3");
exit(EXIT_FAILURE);
}

close(fd1[0]);
close(fd1[1]);
close(fd2[0]);
close(fd2[1]);

waitpid(-1, NULL, 0);
waitpid(-1, NULL, 0);
waitpid(-1, NULL, 0);

return 0;
}

如何在 for 循环中生成 com[n][3]?

最佳答案

“迭代是人,递归是神”——匿名。

我会用递归方法解决这个问题。这是一个非常罕见的场合之一 Three Star programmer几乎是有道理的。 ;)

这完全未经测试,但应该能为您指出正确的方向。

// You'll need to rearrange your command strings into this three dimensional array
// of pointers, but by doing so you allow an arbitrary number of commands, each with
// an arbitrary number of arguments.
int executePipe(char ***commands, int inputfd)
{
// commands is NULL terminated
if (commands[1] == NULL)
{
// If we get here there's no further commands to execute, so run the
// current one, and send its result back.
pid_t pid;
int status;
if ((pid = fork()) == 0)
{
// Set up stdin for this process. Leave stdout alone so output goes to the
// terminal. If you want '>' / '>>' redirection to work, you'd do that here
if (inputfd != -1)
{
dup2(inputfd, STDIN_FILENO);
close(inputfd);
}
execvp(commands[0][0], commands[0]);
perror("execvp");
exit(EXIT_FAILURE);
}
else if (pid < 0)
{
perror("fork");
exit(EXIT_FAILURE);
}
waitpid(pid, &status, 0);
return status;
}
else
{
// Somewhat similar to the above, except we also redirect stdout for the
// next process in the chain
int fds[2];
if (pipe(fds) != 0)
{
perror("pipe");
exit(EXIT_FAILURE);
}
pid_t pid;
int status;
if ((pid = fork()) == 0)
{
// Redirect stdin if needed
if (inputfd != -1)
{
dup2(inputfd, STDIN_FILENO);
close(inputfd);
}
dup2(fds[1], STDOUT_FILENO);
close(fds[1]);
execvp(commands[0][0], commands[0]);
perror("execvp");
exit(EXIT_FAILURE);
}
else if (pid < 0)
{
perror("fork");
exit(EXIT_FAILURE);
}
// This is where we handle piped commands. We've just executed
// commands[0], and we know there's another command in the chain.
// We have everything needed to execute that next command, so call
// ourselves recursively to do the heavy lifting.
status = executePipe(++commands, fds[0]);
// As written, this returns the exit status of the very last command
// in the chain. If you pass &status as the second parameter here
// to waitpid, you'll get the exit status of the first command.
// It is left as an exercise to the reader to figure how to get the
// the complete list of exit statuses
waitpid(pid, NULL, 0);
return status;
}
}

要使用它,首先用 commands 调用它数组按照描述设置,inputfd最初 -1 .

如果你想处理<类型重定向,你可能想检查 inputfd == -1在最顶部,如果需要则进行重定向并替换 inputfd在进入 body 的其余部分之前使用适当的值。

关于c - (在C 中模拟UNIX SHELL)如何在for 循环中实现多个管道?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55362796/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com