How does this bash fork bomb work?


Question

According to Wikipedia, the following is a very elegant bash fork bomb:

:(){ :|:& };:

How does it work?

1
117
6/13/2009 5:45:02 PM

Accepted Answer

Breaking it down, there are three big pieces:

:()      # 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.

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.

A more human-friendly version looks like this:

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.

159
8/28/2009 1:31:40 PM

Short answer:

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.


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