涉及fork()的C程序输出的说明

本文介绍了涉及fork()的C程序输出的说明的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

运行该程序将打印"forked！". 7次.有人可以解释如何分叉！"被打印了7次?

Running this program is printing "forked!" 7 times. Can someone explain how "forked!" is being printed 7 times?

#include<stdio.h> #include<unistd.h> int main(){ fork() && fork() || fork() && fork(); printf("forked!\n"); return 0; }

推荐答案

这里使用了几个概念，第一个概念是知道fork的作用以及在某些情况下返回的结果.不久，当它被调用时，它将创建调用方的重复进程，并在子进程中返回0(对于逻辑表达式，为false)，对于父进程，返回非零(对于逻辑表达式，为true). 实际上，如果发生错误，它可能会返回一个负(非零)值，但是在此我们假定它总是成功.

There're several concepts being used here, first one is knowing what fork does and what it returns in certain circumstances. Shortly, when it gets called, it creates a duplicate process of the caller and returns 0 (false for logical expressions) in child process and non-zero (true for logical expressions) for parent process. Actually, it could return a negative (non-zero) value in case of an error, but here we assume that it always succeeds.

第二个概念是对逻辑表达式(例如&&和||)的短路计算，特别是0 && fork()将 not 调用fork()，因为如果第一个操作数是false(零)，则无需计算第二个.同样，1 || fork()也不会调用fork().

The second concept is short-circuit computation of logical expressions, such as && and ||, specifically, 0 && fork() will not call fork(), because if the first operand is false (zero), then there's no need to compute the second one. Similarly, 1 || fork() will not call fork() neither.

还要注意，在子进程中，表达式的计算在与父进程相同的地方继续进行.

Also note that in child processes the computation of the expression continues at the same point as in the parent process.

此外，请注意，由于优先级高，因此按以下顺序计算表达式:

Also, note that the expression is computed in the following order due to precedence:

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

这些观察结果将引导您找到正确的答案.

These observations should lead you to the correct answer.

考虑fork() && fork()

fork() / \ false true && fork() / \ false true

因此，这里我们创建了三个进程，其中两个返回结果为false，另一个返回true.然后，对于||，我们让所有返回false的进程试图再次运行同一条语句，因此我们将2 * 3 + 1 = 7作为答案.

So here we have three processes created, two of which return false as the result and one returning true. Then for || we have all the processes returning false trying to run the same statement again, so we have 2 * 3 + 1 = 7 as the answer.