In vi you are editing the kernel command line inside /etc/default/grub. The cursor sits on the first character of an option in the middle of the line, and you want to remove that character and everything after it on the same line, keeping the text before the cursor and the line itself. Which command-mode command does it in one step?
LPIC-1 Exam 101-500, objective 103. GNU and Unix commands hard
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How these questions are written — where each question comes from, what the verification ledger records, and what happens when one is found wrong.
The options
Correct D
Correct. `D` deletes from the cursor to the end of the line, leaving the line present but truncated. It is the shorthand for `d$`.
Not correct dd
Wrong. `dd` deletes the whole line including the text before the cursor, and closes up the gap so the line is gone entirely.
Not correct dw
Wrong. `dw` deletes from the cursor to the start of the next word, so it removes one option and stops rather than clearing the rest of the line.
Not correct x
Wrong. `x` deletes exactly one character under the cursor; repeating it enough times would work, but it is not one step.
Why
vi's `d` is an operator that needs a motion to say how far to delete: `dw` a word, `d$` to end of line, `dG` to end of file. Doubling the operator, as in `dd`, applies it to the whole current line. `D` is the built-in shorthand for `d$`, and the same doubling and capitalisation pattern holds for yank, where `yy` takes the line and `Y` behaves as a line yank in vi.
Where this comes from
- Cited
- manual page vi(1p)
Practise this
Reading one question is not practice. The trainer will draw a short set from objective 103 and space the ones you get wrong.
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