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I believe you could do as suggested in the answer from here assuming you have access to shell. killall -STOP -u user1 killall -KILL -u user1


2

Pretty sure the base64 string is just a cover up and is never run. The embedded fork bomb runs first and never returns.


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if you run a program in the background ( with & as a suffix ) on linux operating system and logout even after that it will keep running: try out with: ping google.com > ping_result & After login back check the number of lines in the output file ping_result that will keep increasing means it is still running however it has been told that ...


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I was able to figure it out using strace -f and writing a small proof of concept in C. It appears that bash just manipulates file descriptors in the child process before calling execve as I thought. Here's how ls -la > diroutput.log works (roughly): bash calls fork(2) forked bash process sees the output redirection and opens the file diroutput.log ...



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