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Exam 1z0-808 topic 1 question 84 discussion

Actual exam question from Oracle's 1z0-808
Question #: 84
Topic #: 1
[All 1z0-808 Questions]

Given:

What is the result?

  • A. Compilation fails.
  • B. 11
  • C. 8
  • D. 9
  • E. 10
Show Suggested Answer Hide Answer
Suggested Answer: B 🗳️
by JoseCG at Oct. 25, 2019, 12:05 a.m.

Comments

Chosen Answer:
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Zafar_Nasim
Highly Voted 4 years, 7 months ago
D is correct because trim() only removes leading and lagging spaces.
upvoted 9 times
haisaco
2 years ago
No trim() just remove space on first and last string
upvoted 4 times
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carloswork
Highly Voted 2 years, 2 months ago
Selected Answer: B
Answer is B. String s = "Java SE 8 1"; int len = s.trim().length(); System.out.print(len);
upvoted 6 times
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MPignaProTech
Most Recent 3 months ago
Selected Answer: B
trim delete spaces at the begining and the end
upvoted 1 times
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Jess1985B
1 year, 1 month ago
Answer is B - 11. Just calculate spaces before and after string.
upvoted 2 times
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dsms
1 year, 5 months ago
Answer is: 11
upvoted 1 times
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haisaco
2 years ago
Selected Answer: B
Answer is B.
upvoted 2 times
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tawa_z58
2 years, 1 month ago
Answer is B trim() removes only begining and end white spaces of a string
upvoted 5 times
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iSnover
2 years, 2 months ago
Selected Answer: B
See the code: public static void main (String[] args) { String s = "Java SE 8 1"; int len = s.trim().length(); System.out.print(len); }
upvoted 4 times
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iSnover
2 years, 3 months ago
Selected Answer: D
D is correct, because trim remove only spaces and remember, can't be 8 because is true that Java indexes form 0, but not in case of count de size of String. I tested
upvoted 1 times
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archer1903
2 years, 7 months ago
Selected Answer: D
Tested D. 9
upvoted 1 times
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lilz
3 years, 10 months ago
String s = "Java Duke"; int len = s.trim().length(); System.out.println(len); D
upvoted 1 times
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JoseCG
5 years, 2 months ago
Correct.
upvoted 4 times
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Community vote distribution
A (35%)
C (25%)
B (20%)
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