同时执行多个进程

I am working on a home made shell (very simple shell). I have decided to take the route of using execvp as my path is not a changeable element for my shell. I am running into an issue with coming up with the logic on how to fork and exec multiple processes at once.

我正在做一个自制的外壳（非常简单的外壳）。我决定采用execvp的路线，因为我的路径不是我的shell的可变元素。我遇到了一个关于如何一次分叉和执行多个进程的逻辑的问题。

My program should work with a command as such:

我的程序应该使用如下命令：

ls ; echo hello ; cat shell.c

Where each ";" indicates that we would like to run these processes at once simultaneously. So on our terminal output we should get a mix of these commands working at once.

每个“;”表示我们希望同时运行这些进程。所以在我们的终端输出上，我们应该同时使用这些命令的混合。

To elaborate I'd like to explain how my program works:

详细说明我想解释一下我的程序是如何工作的：

A. Intake full command line into char array with a grab line function
B. Split the char array received from the last function by delimiters and place into an array of char arrays (pointer to pointer).
C. If one of the elements in our array of char arrays is ";" we can assume that multi commands are necessary. This is where I have trouble.

I have gotten as far as to know exactly how many processes I need to fork and such, but I cannot seem to wrap my head around how to pass all of these functions plus their arguments to the execvp function at once. Should I use a temp array? I know this shouldn't be this complicated but for some reason I cannot figure it out. I'm submitting my launch function below, which intakes an array of char arrays and executes accordingly based on my "multiCommand" variable which is set when multi commands are needed (by my split line function)

我已经知道到底需要多少进程来分叉等等，但我似乎无法理解如何将所有这些函数及其参数同时传递给execvp函数。我应该使用临时数组吗？我知道这不应该是这么复杂，但由于某种原因，我无法弄明白。我正在下面提交我的启动函数，它会引入一个char数组并根据我的“multiCommand”变量执行相应的操作，该变量在需要多个命令时设置（通过我的分割线函数）

int launch(char **args){

    pid_t pid;
    int status;
    int i = 0;

    if(strcmp(args[0], "quit") == 0){
        exit(EXIT_SUCCESS);
    }

    if(strcmp(args[0], ";") != 0){
        printf("Essential Command Found : %s\n", args[0]);
        numFork++;
    }


    if(multiCommand == 1){
        //Handle Multicommands here
        printf("Multi Commands Handling Here\n");

        for(; i < elements - 1; i++){
            if(strcmp(args[i], ";") == 0){
                if((i + 1) < elements){
                    printf("Essential Command Found : %s\n", args[i + 1]);
                    numFork++;
                }
            }
        }

        //This is where I need to figure out what to do

        printf("Fork: %d times\n", numFork);


    }else if (multiCommand == 0){
        pid = fork();
        if(pid == 0){
            execvp(args[0], args);

        }else{
            wait(&status);
        }
    }

    multiCommand = 0;   
    elements = 0;

    return 1;
}

1 个解决方案

#1

The general idea would be to have a for loop over the different commands and fork each of them.

一般的想法是对不同的命令进行for循环并分叉它们中的每一个。

E.g.

例如。

for(int i = 0; i < commandCount; i++) {
    int pid = fork();
    if(pid == 0) { //this is the child (don't forget to check for errors and what-not)
        execCommand(all, of, the, info, needed);
    }
}

You can easily get the different commands using strtok().
Here's an example:

您可以使用strtok（）轻松获取不同的命令。这是一个例子：

#include <string.h>
#include <stdio.h>
int main() {
    char input[] = "abc;def;ghi";
    char *token = strtok(input, ";");
    while(token != NULL) {
        printf("%s\n", token);
        token = strtok(NULL, ";");
    }
    return 0;
}

Output:

输出：

abc
def
ghi

The final function will look something like this:

最终的功能看起来像这样：

char *token = strtok(input, ";");
while(token != NULL) {
    int pid = fork();
    if(pid == 0) {
        //token is one command
        //parse the different parts here
        execCommand(args, to, exec);
    }
    token = strtok(NULL, ";");
}

#1