What is true about the dmesg command? (Choose two.)
What is true about the dmesg command? (Choose two.)
The dmesg command displays the content of the Linux kernel's ring buffer, which contains system messages and diagnostic information. Additionally, the ring buffer has a limited size, and when it becomes full, the oldest messages are overwritten by newer information. Therefore, dmesg might not display older information if it has been replaced by newer entries.
c , e are correct. I just took the exam and got it right
How can you know if you got it right or not? you can just get the score and the result. So you can't know which answer you got right or wrong....
You can figure it out if you get a 100 on a particular section and you know the question is from that section.
You have no idea of knowing if you got it right or wrong! The test does not provide you with details on which questions you got right or wrong.
What's he doing back here if he took the exam, must have not passed.
C AND E is correct.
From Man Page for C: dmesg is used to examine or control the kernel ring buffer. The default action is to display all messages from the kernel ring buffer. From Stackexchange question for E (https://unix.stackexchange.com/questions/181067/how-to-read-dmesg-from-previous-session-dmesg-0): To clear up a fundamental misconception, dmesg does not read from /var/log/dmesg. It reads directly from the kernel ring buffer and gives you the most recent N messages.
What do these symptoms mean? ’ They are in several of the questions and answers but I have not been able to find a meaning for them or how they fit into the syntax. Can someone please help I will be taking this exam in abut 10 days and I really need to help with this!?
It's a ’ (RIGHT SINGLE QUOTATION MARK - U+2019) character which has been encoded as CP-1252 instead of UTF-8. If you check the encodings table, then you see that this character is in UTF-8 composed of bytes 0xE2, 0x80 and 0x99. If you check the CP-1252 code page layout, then you'll see that each of those bytes stand for the individual characters â, € and ™.
the "dmesg" command prints the kernel ring buffer and new messages are overwritten
C. It displays the content of the Linux kernel's ring buffer. E. It might not display older information because it was overwritten by newer information.
CE is correct