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Linux Foundation LFCS Exam - Topic 7 Question 52 Discussion

What is the difference between the commands test -e path and test -f path?
A) They are equivalent options with the same behaviour.
B) The -f option tests for a regular file. The -e option tests for an empty file.
C) Both options check the existence of the path. The -f option also confirms that it is a regular file.
D) The -f option tests for a regular file. The -e option tests for an executable file.

Linux Foundation LFCS Exam - 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?

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Suggested Answer: E

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

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Angella
10 months ago
So basically, -f is a stricter check? Makes sense!
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Natalie
11 months ago
Nope, -e just checks if the path exists, not empty or executable.
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Shaquana
11 months ago
Wait, I thought -e was for executable files?
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Wendell
11 months ago
Totally agree, -f is more specific!
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Zack
11 months ago
The -f option checks for regular files, while -e just checks existence.
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Corrie
11 months ago
I thought -e was for executable files, but now I'm confused because I remember it being about existence too.
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Sarah
11 months ago
I believe both options check for existence, but -f specifically confirms it's a regular file. That sounds right, doesn't it?
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Precious
11 months ago
I remember practicing something similar where -f was about regular files, but I can't recall exactly what -e checks for.
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Robt
11 months ago
I think the -e option just checks if the file exists, but I'm not entirely sure what -f does.
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Gilma
12 months ago
I'm pretty sure the right answer has to do with security and licensing requirements for Flow. I just need to make sure I understand the details.
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Lashandra
12 months ago
The key is focusing on the specific details in the question. A voucher has to include all three elements - the purchase order, vendor invoice, and receiving reports. I think A is the only option that matches that.
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Lacresha
12 months ago
Hmm, this one seems tricky. I'll need to think through how Selenium fits into the overall test automation architecture.
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Margret
1 year ago
Wait, is this a trick question? I feel like I'm being Schrodinger'd here. I'm just going to go with C and hope for the best.
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Laticia
1 year ago
User 3: So, the -f option also confirms it's a regular file. I'll go with C too.
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Christa
1 year ago
User 2: Yeah, that sounds right. The -e option just checks for existence.
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Blair
1 year ago
User 1: I think the -f option checks for a regular file.
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Hermila
1 year ago
D? Really? That's just silly. I mean, who confuses regular files with executable files these days? Option C is the way to go, folks.
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Rikki
1 year 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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Denny
1 year ago
C is the clear winner here.
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Leota
1 year ago
I almost went with A, but then I realized that the options have different behaviors.
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Lavina
1 year 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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Bernardo
1 year ago
User 2: Yeah, both options check the existence of the path, but -f also confirms it's a regular file.
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Lawrence
1 year ago
User 1: I think C is the correct option.
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Mona
1 year 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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Danilo
1 year 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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Lezlie
1 year 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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Mona
1 year ago
I think the difference is that -f checks for a regular file, while -e checks for an empty file.
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