LPIC-1 Shells, Scripting and Data Management Practice Question
A script needs to remove all trailing whitespace (spaces and tabs) from each line of a file. Which sed command will accomplish this?
⚠ Common exam trap
It's easy for candidates to confuse the quantifiers '*' (zero or more) and '+' (one or more) or forget that sed's default BRE mode requires escaping the '+' quantifier, leading them to choose Option A which would not work as intended.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
sed 's/[ \t]*$//'
The sed command 's/[ \t]*$//' uses a regular expression that matches zero or more spaces or tabs (the character class [ \t] followed by *) anchored at the end of the line ($) and replaces them with nothing, effectively removing all trailing whitespace. The * quantifier ensures that both single and multiple trailing whitespace characters are removed, and the $ anchor restricts the match to the end of the line only.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
sed 's/[[:space:]]+$//'
Why it's wrong here
Needs -E for extended regex; in basic + is literal.
- ✗
sed 's/[ \t]*//g'
Why it's wrong here
Removes all whitespace globally.
- ✗
sed 's/^[ \t]*//'
Why it's wrong here
Removes leading whitespace.
- ✓
sed 's/[ \t]*$//'
Why this is correct
Matches trailing spaces/tabs and replaces with nothing.
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