程序流程，带有逻辑的 fork()

Question

我正在尝试理解这个简单的 C fork() 代码的流程：

fork() && fork() || fork();
fork();
printf("forked ");

我的输出（g++）是：

forked forked forked forked forked forked forked forked forked 

跳过它并不能真正帮助我理解它。

Answer 1

fork() && fork() || fork();
^^^^^^^^^^^^^^^^
Parent forks a child with return value 0 for the child process. 
Since it is logical AND operator, the short-circuit evaluation applies, 
so the child doesn’t go further. Parent goes on forking again(right-hand-side of AND). 
There are 3 processes being composed of 2 children and 1 parent.



fork() && fork() || fork();
                    ^^^^^^
                    Again parent forks one more and since OR operator
                    doesn’t apply short-circuit for latter child
                    (due to return type which is 0 and left-hand-side of OR op.), 
                    the latter child forks as well. There are totally 5 processes.

 fork()[fourth]; doubles prior number of processes 5*2 = 10 processes totally exist.

您的输出很可能被缓冲了。试试fprintf(stderr, "forked ");、fflush(stdout) 或printf("forked\n”);。

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我认为下面的图更有帮助。