叉混淆（） [英] Confused with fork()

问题描述

我有困难的时候理解不同情景下fork（）的命令做了什么。下面是我的一些书样本code：

  INT的main（）{
    int类型的= 12;
    INT B = 9;
    INT FID = fork（）的;    如果（FID == 0）{
        一个++;
    }
    其他{
        等待（NULL）;
        B =  -  5;
    }    的printf（程序退出时=％D，B =％d个\\ n，A，B）;
    返回0;
}
 

有人能走我通过什么叉（）命令在这种情况下，这样做或许能给一些例子，以澄清？

解决方案

  [主]
                 一个= 12
                 B = 9                 叉子（）
                   |
                   |
    + -------------- -------------- + +
    | |
    | |
[家长] [儿童]FID = 1234 FID = 0
等待（NULL）一++
   ...的printf（A = 13，B = 9）;
   ...返回0
   ...＆LT;退出＆GT;
B =  -  5
输出（α= 12时，b = 4）;
返回0
＆LT;退出＆GT;
 

在叉（）执行有计划的两个副本。每个进程都有自己的变量副本，所以现在有两个 A 的两个 B 的，等两个方案之间的唯一区别是叉从返回的值（）：在子进程叉（）返回0;在父进程返回子的PID。

子进程增量 A ，打印 A 和 B ，并退出

父进程首先等待子进程终止。孩子已经完成了它继续，从减去5 B ，打印出只有在 A 和 b ，然后退出。

的等待（NULL）确保子进程的打印始终到来之前父母的，所以你总是会得到一个可靠的订单相同的输出。没有它，你就无法依靠两个打印输出的数量级上。他们将是相同的消息，只是在未predictable秩序。

I am having a difficult time understanding what the fork() command does under different scenarios. Here is some sample code from my book:

int main() {
    int a = 12;
    int b = 9;
    int fid = fork();

    if (fid == 0) {
        a++;
    }
    else {
        wait(NULL);
        b = b - 5;
    }

    printf("program exit. a = %d, b = %d\n", a, b);
    return 0;
}

Can someone walk me through what the fork() command is doing in this case and perhaps give some more examples to clarify?

解决方案

                 [main]
                 a = 12
                 b = 9

                 fork()
                   |
                   |
    +--------------+--------------+
    |                             |  
    |                             |
[parent]                       [child]

fid = 1234                     fid = 0
wait(NULL)                     a++
   ...                         printf("a = 13, b = 9");
   ...                         return 0
   ...                         <exit>
b = b - 5
printf("a = 12, b = 4");
return 0
<exit>

After fork() executes there are two copies of the program. Each process gets its own copies of the variables, so there are now two a's, two b's, etc. The only difference between the two programs is the value returned from fork(): in the child process fork() returns 0; in the parent process it returns the PID of the child.

The child process increments a, prints the values of a and b, and exits.

The parent process first waits for the child process to terminate. Only after the child has finished does it continue on, subtracting 5 from b, printing out a and b, and then exiting.

The wait(NULL) ensures that the child process's printout always comes before the parent's, so you will always get the same output in a reliable order. Without it you wouldn't be able to depend on the order of the two printouts. They would be the same messages, just in an unpredictable order.

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