LPIC-1 read command Practice Question
Exhibit
Consider the following script named 'parse.sh': #!/bin/bash while IFS=',' read -r field1 field2 remainder; do echo "$field1 $field2" done < input.txt Assume input.txt contains: a,b,c d,e,f
Refer to the exhibit. What will be the output when this script is executed?
⚠ Common exam trap
The trap is that `read` assigns the remainder of the line to the last variable when there are more words than variables. Candidates often think each variable gets exactly one word, but actually the last variable absorbs all remaining words.
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
✓
a b c\nd e f
The script reads two lines from standard input. The first `read a b` reads the first line 'a b c'. Since there are three words but only two variables, the last variable `b` captures the remainder of the line, so a='a' and b='b c'. The second `read d e` reads the second line 'd e f', so d='d' and e='e f'. The `echo -e` command outputs the values with an escaped newline (`\n`) between them, producing two lines: 'a b c' and 'd e f', represented as 'a b c\nd e f'.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
a b\nd e
Why it's wrong here
Incorrect. This output would occur if `read` assigned only the first word to each variable and discarded the rest, but the last variable actually captures the remainder.
- ✓
a b c\nd e f
Why this is correct
Correct. Because `b` and `e` capture the remaining words from their respective lines, the output includes 'b c' and 'e f'.
- ✗
a b
Why it's wrong here
Incorrect. This output would result if the script only printed the first variable from each line, ignoring the rest.
- ✗
The script will error because of incorrect read syntax.
Why it's wrong here
Incorrect. The syntax is valid; `read` does not error when there are more words than variables.
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