Question

我开始在UNIX中编写shell来练习API调用，例如fork（）dup2（），read（）和wait（）。目前，我的shell打开并运行正常。当我键入要运行的命令（例如ls -a）时，它会正确解析此命令并执行它。问题是，主循环提前终止，在单个命令后退出shell。我需要循环继续运行，直到从stdin读取'exit'。这是我目前的代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <fcntl.h>
#include <sys/wait.h>
#include <unistd.h>
#include <signal.h>
#include <unistd.h>
static const char prompt[] = "myshell> ";
static const char sep[] = " \t\n\r";
int main()
{
    int ac; // arg count
    char *av[10]; //argument vector
    int tty = open("/dev/tty", O_RDWR); // open tty for read/write
    int pid; // process id
    int status; // child process exit status
    int w;
    void (*istat)(int), (*qstat)(int);

    if (tty == -1)
    {
        fprintf(stderr, "can't open /dev/tty\n");
        exit(EXIT_FAILURE);
    }
    while (1)
    {
        char *arg, line[256]; // buffer to hold line of input
        int i;
        // prompt and read
        write(tty, prompt, sizeof(prompt) - 1);
        i = read(tty, line, sizeof(line));
        if (i <= 0)
            break;
        line[i] = '\0';
        // tokenize the line into av[]
        ac = 0;
        for (arg = strtok(line, sep); arg && ac < 10; arg = strtok(NULL, sep))
            av[ac++] = arg;

        if (ac > 0 && strcmp(av[0], "exit") == 0)
            break;

        if ((pid = fork()) == 0)
        {
            // this is the forked child process that is a copy of the running program
            dup2(tty, 0); // stdin from tty
            dup2(tty, 1); // stdout to tty
            dup2(tty, 2); // stderr to tty
            close(tty);
            // last argument must be NULL for execvp()
            av[ac] = NULL;
            // execute program av[0] with arguments av[0]... replacing this program
            execvp(av[0], av);
            fprintf(stderr, "can't execute %s\n", av[0]);
            exit(EXIT_FAILURE);
        }
        close(tty);
        // disable interrupt (^C and kill -TERM) and kill -QUIT
        istat = signal(SIGINT, SIG_IGN);
        qstat = signal(SIGQUIT, SIG_IGN);
        // wait until forked child process terminated, get its exit status
        while ((w = wait(&status)) != pid && w != -1)
            continue;
        if (w == -1)
            status = -1;

    }

    // restore interrupt and quit signals
    signal(SIGINT, istat);
    signal(SIGQUIT, qstat);
    exit(EXIT_SUCCESS);
}

我尝试在退出成功之上移动这些线（因此它们在循环之外和主要内部）

    close(tty);
    // disable interrupt (^C and kill -TERM) and kill -QUIT
    istat = signal(SIGINT, SIG_IGN);
    qstat = signal(SIGQUIT, SIG_IGN);
    // wait until forked child process terminated, get its exit status
    while ((w = wait(&status)) != pid && w != -1)
        continue;
    if (w == -1)
        status = -1;

}

// restore interrupt and quit signals
    signal(SIGINT, istat);
    signal(SIGQUIT, qstat);
    exit(EXIT_SUCCESS);
}

这样做是正确循环shell，但是在执行命令的输出的最后一行上方打印提示，因为程序在打印提示并等待输入之前没有等待子进程结束。我尝试移动一些线路，但每次执行一个命令后它最终都会终止shell

Answer 1

您不想在主循环中调用close(tty)。这样做会导致下一个read(tty,...失败，退出shell。

此外，如果你真的想要禁用SIGQUIT / SIGQUIT，你应该在循环中对称地恢复它们。

Answer 2

以下发布的代码：

跳过信号的处理，你可能想要重新加入。
干净地编译
正确检查并记录任何错误
执行所需功能，但禁用某些信号
继续执行循环，直到用户输入＆＃39;退出＆＃39;

现在，建议的代码：

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

// added next statement for 'wait()' and 'waitpid()'
#include <sys/types.h>

#include <sys/wait.h>
#include <unistd.h>
//#include <signal.h>
#include <unistd.h>

static const char prompt[] = "myshell> ";
static const char sep[] = " \t\n\r";


int main( void )
{
    int ac; // arg count
    char *av[10]; //argument vector

    pid_t pid; // process id
    int status; // child process exit status


    int tty = open("/dev/tty", O_RDWR); // open tty for read/write
    if (tty == -1)
    {
        perror( "open for /dev/tty failed");
        exit(EXIT_FAILURE);
    }

    while (1)
    {
        char *arg;
        char line[256]; // buffer to hold line of input

        // prompt and read
        write(tty, prompt, sizeof(prompt) - 1);
        ssize_t i = read(tty, line, sizeof(line));

        if (i == 0)
            break;

        if (i < 0)
        {
            perror( "read failed" );
            exit( EXIT_FAILURE );
        }

        line[i] = '\0';

        // tokenize the line into av[]
        ac = 0;
        for (arg = strtok(line, sep); arg && ac < 10; arg = strtok(NULL, sep))
            av[ac++] = arg;

        // last argument must be NULL for execvp()
        av[ac] = NULL

        if (ac > 0 && strcmp( av[0], "exit" ) == 0)
            break;

        pid = fork();
        switch( pid )
        {
        case -1:
            perror( "fork failed" );
            exit( EXIT_FAILURE );
            break;

        case 0:  // child process
            // this is the forked child process that is a copy of the running program
            dup2(tty, 0); // stdin from tty
            dup2(tty, 1); // stdout to tty
            dup2(tty, 2); // stderr to tty
            close(tty);
            ;

            // execute program av[0] with arguments av[0]... replacing this program
            execvp(av[0], av);
            perror( "execvp failed" );
            exit(EXIT_FAILURE);
            break;

        default: // parent process
            // wait until forked child process terminated, get its exit status
            waitpid( pid, &status, 0 );
            break;
        }
    }

    return 0;
}

UNIX：使用fork（）和dup2（）在循环中运行进程会提前结束循环

2 个答案: