这个 bash fork 炸弹是如何工作的？
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How does this bash fork bomb work?
提问by Lajos Nagy
According to Wikipedia, the following is a very elegant bash fork bomb:
根据维基百科，以下是一个非常优雅的 bash fork 炸弹：
:(){ :|:& };:
How does it work?
它是如何工作的？
回答by John Feminella
Breaking it down, there are three big pieces:
将其分解为三大块：
:() # Defines a function, ":". It takes no arguments.
{ ... }; # The body of the function.
: # Invoke the function ":" that was just defined.
Inside the body, the function is invoked twice and the pipeline is backgrounded; each successive invocation on the processes spawns even more calls to ":". This leads rapidly to an explosive consumption in system resources, grinding things to a halt.
在主体内部，函数被调用两次，管道被后台处理；对进程的每次连续调用都会产生更多对“:”的调用。这会迅速导致系统资源的爆炸性消耗，使事情陷入停顿。
Note that invoking it once, infinitely recursing, wouldn't be good enough, since that would just lead to a stack overflow on the original process, which is messy but can be dealt with.
请注意，调用它一次，无限递归，还不够好，因为这只会导致原始进程的堆栈溢出，这很混乱但可以处理。
A more human-friendly version looks like this:
更人性化的版本如下所示：
kablammo() { # Declaration
kablammo | kablammo& # The problematic body.
}; kablammo # End function definition; invoke function.
Edit:William's comment below was a better wording of what I said above, so I've edited to incorporate that suggestion.
编辑：威廉在下面的评论是对我上面所说的更好的措辞，因此我进行了编辑以纳入该建议。
回答by Talljoe
Short answer:
简短的回答：
The colon (":") becomes a function, so you are running the function piped to the function and putting it in the backgroun which means for every invocation of the function 2 copies of the function are invoked. Recursion takes hold.
冒号 (":") 成为一个函数，因此您正在运行通过管道传输到该函数的函数并将其放入后台，这意味着每次调用该函数都会调用该函数的 2 个副本。递归成立。
