Question

我正在编写这个程序，创建一个父母和一个孩子，我必须验证一个用户名。但似乎它没有工作，因为我有一个全局变量集，所以我可以直接跳过一些代码块到我想要评估用户名的部分。但似乎我拥有的变量logEval并没有保留我在第一个子节点中给出的值，所以它在父节点内返回0。

这是代码：

var timeout;
$("#iframe").attr("src", "url-cross-domain.html").on("load", function() {
    clearTimeout(timeout);
    console.log("loaded !");
}):
timeout = setTimeout(function() {
    $("#iframe").off("load").remove();
    console.error("iframe loading failed");
}, 60000);

第二次编辑：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <unistd.h>
#include <fcntl.h>
//sys variables

char Input[100], commandLogin[100], Output[100], Username[50], possibleUsername[50];
int readDescriptor, logEval;
pid_t childPid;

//

int commandCatcher(){

    //define socket
    int socketOne[2], socketTwo[2];

    if(socketpair(AF_UNIX, SOCK_STREAM, 0, socketOne) < 0){
        perror("socket - err");
        exit(0);
    }

    if(logEval == 0)
    switch(fork()){//first child
        case -1:
            perror("fork - err");
            exit(1);

        case 0:
            childPid=getpid();
            readDescriptor = read(socketOne[0], commandLogin, sizeof(commandLogin));

            if (strcmp(commandLogin, "login") == 0 ){
                logEval = logEval + 1;
                printf("Log eval from first:%d\n", logEval);
                write(socketOne[0], "ok", strlen("ok") + 1);
                //something to do here so that the child knows it was given the username

                exit(1);
            }

            else {
                printf("Try again.\n");
                write(socketOne[0], "none", strlen("none")+1);
                exit(1);
            }
    }

    if(logEval == 1){

        switch(fork()){//verify username
            case -1:
                perror("fork - err");
                exit(2);

            case 0:
                readDescriptor = read(socketTwo[0], possibleUsername, sizeof(possibleUsername));
                printf("Username from second child:%s\n", possibleUsername);
                printf("logEval is : %d", logEval);
        }
        exit(2);
    }

    //parent

    //getting initial value
    scanf("%s", Input);
 //writing initial value
    write(socketOne[1], Input, strlen(Input)+1);

    readDescriptor = read(socketOne[1], Output, sizeof(Output));
    //printf("output    %s\n", Output);

//verify if the command was correctly given
    if(strcmp(Output, "ok") == 0) {
        printf("Command was accepted. Insert your username: %d\n", logEval);

        if(socketpair(AF_UNIX, SOCK_STREAM, 0, socketTwo) < 0){//second socket
            perror("sockettwo - err");
            exit(2);
        }

        scanf("%s", Username);
        write(socketTwo[1], Username, strlen(Username)+1);
        printf("%s\n", Username);

        printf("%d\n", logEval);

        commandCatcher();
    }
    else if(strcmp(Output, "none") == 0) {
        printf("Command was denied. Please try again:\n");

        commandCatcher();//recursive call
    }

wait(&childPid);
printf("execution finished");
}

int main(){
    printf("Welcome to Sys v1.0. To start off, please insert your command. \n");
    commandCatcher();
    return 0;
}

Answer 1

当您fork()进程时，它的孩子无法访问父母的地址空间。它将拥有它的完整副本，因此在fork()之前设置的任何内容在子项中都是相同的，但如果您更新子进程内的任何内容，则父进程将不会看到它。您需要为此使用某种形式的进程间通信（IPC），例如共享内存或管道。

Linux变量不维护

1 个答案: