Linux Foundation Exam LFCS Topic 7 Question 52 Discussion

Actual exam question for Linux Foundation's LFCS exam

Question #: 52
Topic #: 7

[All LFCS Questions]

What is the difference between the commands test -e path and test -f path?

AThey are equivalent options with the same behaviour.

BThe -f option tests for a regular file. The -e option tests for an empty file.

CBoth options check the existence of the path. The -f option also confirms that it is a regular file.

DThe -f option tests for a regular file. The -e option tests for an executable file.

Show Suggested Answer

Suggested Answer: E

by Filiberto at Jan 08, 2023, 01:02 AM

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2 days ago

I almost went with A, but then I realized that the options have different behaviors. C is the clear winner here.

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11 days ago

I was a bit confused between the -e and -f options, but after reading the question closely, I think C is the way to go.

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12 days ago

Oh, I see. So -f not only checks for existence but also confirms it's a regular file. That makes sense.

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16 days ago

Option C seems to be the correct answer. Both options check for the existence of the path, but -f also confirms that it is a regular file.

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19 days ago

Actually, -f checks for a regular file, and -e checks for the existence of the path. So, -f is more specific.

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21 days ago

I think the difference is that -f checks for a regular file, while -e checks for an empty file.

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