在以下程序中将产生多少进程

问题描述

int main()
{
fork();
fork() && fork() || fork();
fork();

printf("forked\n");
return 0;
}

当我们调用 fork 函数时，父进程得到一个 非零 pid 而子进程得到一个 0 作为回报基于这个逻辑，在第二个语句中，我们将不得不应用短路(afaik)的原则....第一次调用后会有2个进程，

When we call fork function , the parent gets a non-zero pid while the child gets a 0 in returnBased on this Logic, in the second statement we will have to apply the principle of shortcircuiting(afaik)....After the 1st call there will be 2 process,

在第二行之后，有 8 个进程，[父进程在 (fork() && fork()) 中分叉了两次，第二个子进程也因"fork()||fork()"]

After the second line, 8 processes,[A parent gets forked two times in (fork() && fork()), and the second child also gets twice due to the "fork()||fork()"]

最后是 16(根据我的推理)

And finally 16(According to my reasoning)

请告诉我这是正确的还是涉及其他逻辑

Please let me know if this is correct or some other logic is involved

推荐答案

要计算 (fork() && fork() || fork()) 之后的进程数，只需请记住:

To calculate the number of process after (fork() && fork() || fork()), just remember that :

• In (&&) 逻辑运算符:右侧仅在左侧为非零时计算

• In (&&) logical operator : the right side is evaluated only when the left side is NON ZERO

In (||) 逻辑运算符:右侧仅在左侧为零

In (||) logical operator : the right side is evaluated only when the left side is ZERO

使用运算符优先级，我们可以这样写:

And with operators precedence we wan write this like so:

(fork() && fork()) ||fork()

还要记住 fork 返回 NON ZERO 给 parent 和 ZERO 给 child

Remember also that fork return NON ZERO to the parent and ZERO to the child

为了简化说明，我们重命名:

To simplify the explanation, we rename :

fork() &&fork() 操作 A

和最后的 fork() 操作 B，所以前一行相当于:

and the last fork() Operation B, so the precedent line is equivalent to:

(fork() && fork()) ||fork() => A ||乙

第一行(fork):

---> 2 个进程(Father 和 Child1)

第二行:

• 操作:

第一个叉 =>

父亲会给孩子=> 父亲(Child2的PID)和Child2(零)Child1 将给一个孩子 => Child1(Child3 的 PID) 和 Child3(ZERO)

Father will give a child => Father(PID of Child2) and Child2(ZERO)Child1 will give a child => Child1(PID of Child3) and Child3(ZERO)

我们有4个进程:Father(Child2的PID)、Child2(零)、Child1(Child3的PID)和Child3(零)

(&& fork()) 将仅对返回非零 => Father 和 Child1 的最后操作执行

The (&& fork()) will be executed only for last operations that returns NON ZERO => Father and Child1

父亲愿意给孩子=> 父亲(Child4的PID)和Child4(零)Child1 将给一个孩子 => Child1(Child5 的 PID) 和 Child5(ZERO)

Father wille give a child => Father(PID of Child4) and Child4(ZERO)Child1 will give a child => Child1(PID of Child5) and Child5(ZERO)

总结一下:

我们有6个流程:

父亲(Child4 的PID)、Child4(零)、Child1(Child5 的PID)、Child5(零)、Child2(零)和Child3(零)

• B 操作:

只对最后一个返回零的命令执行 => 从一个操作中返回零的进程，关注的进程是:

Is only executed for last commands that returns ZERO => process that returns ZERO from A Operation, Concerned Process are:

Child4(零)、Child5(零)、Child2(零)和 Child3(零)

当分叉这 4 个进程时，我们以 4 个新进程结束 => 第二行之后的进程总数 = 10

When forking this 4 process we end with 4 new process => Total number of process after second line = 10

第三行:这只是一个简单的叉子

Third Line :It's just a simple fork

=> 进程总数= 20

为了证明:使用这个 (fork_quiz.c)

To demonstrate that: use this (fork_quiz.c)

#include <unistd.h>
#include <stdio.h>

int main(int argc, char **argv)
{
        fork();
        fork() && fork() || fork();
        fork();

        sleep(10);

        return 0;
}

并编译它:

gcc -Wall fork_quiz.c -o fork_quiz

然后像这样运行:

toc@UnixServer:~$ ./fork_quiz & (sleep 1; ps -o "%P%p%c")
[1] 15455
 PPID   PID COMMAND
15046 15047 bash
15047 15455 fork_quiz
15047 15456 bash
15455 15457 fork_quiz
15455 15458 fork_quiz
15455 15459 fork_quiz
15455 15460 fork_quiz
15457 15462 fork_quiz
15457 15463 fork_quiz
15457 15464 fork_quiz
15458 15465 fork_quiz
15458 15466 fork_quiz
15459 15467 fork_quiz
15459 15468 fork_quiz
15465 15469 fork_quiz
15467 15470 fork_quiz
15463 15471 fork_quiz
15463 15472 fork_quiz
15462 15473 fork_quiz
15462 15474 fork_quiz
15473 15475 fork_quiz
15471 15476 fork_quiz
15456 15477 ps

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