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Consider the following code:
What change should be made to ensure that it correctly displays the value of name in all uppercase letters?
Suggested Answer: B
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35 comments · newest first
JA
Jade
11 months ago
D is just wrong, it's "string" not "sting"!
WI
Willie
11 months ago
Wait, is it really "new sting"? That seems off.
TA
Tammara
11 months ago
C won't work, NAME isn't defined.
AL
Alishia
12 months ago
I think B is incorrect, prototype isn't needed here.
MA
Malcom
12 months ago
A is definitely the right choice!
WI
Willis
12 months ago
Option C looks off to me; I don't recall using uppercase like that in document.write.
RE
Refugia
1 year ago
I remember a similar question where we had to manipulate strings, and I think changing line 2 is the right approach.
SH
Shawn
1 year ago
I'm not entirely sure, but option B seems wrong because prototype methods are not called like that.
MI
Mitsue
1 year ago
I think option A makes the most sense since it directly modifies the variable to be uppercase.
AR
Arlene
1 year ago
This looks like a good opportunity to demonstrate my understanding of string manipulation in JavaScript. I think option A is the correct answer, as it's the most straightforward way to convert the `name` variable to uppercase.
MA
Marshall
1 year ago
Okay, let's see. The issue is that the `name` variable is not being converted to uppercase before it's displayed. I think option A is the right solution - we just need to call `toUpperCase()` on the `name` variable.
MY
My
1 year ago
I'm a bit confused by this question. The code seems to be displaying the value of `name`, but it's not clear why it's not in uppercase. I'll need to think this through carefully.
MA
Maybelle
1 year ago
Hmm, this looks like a straightforward question about string manipulation. I think the answer is A, since the `toUpperCase()` method is the standard way to convert a string to uppercase.
TA
Tamala
1 year ago
I'm not sure about this one. The code seems simple enough, but I'm not confident I understand the difference between the options. I'll need to review the string methods in JavaScript to make sure I'm choosing the right solution.
ZA
Zack
1 year ago
Okay, I've got this. The key is to leverage existing resources like CVDs, connect all assets on the network, and use the data to drive business outcomes. That's the most comprehensive approach to engaging customers and developing IoT opportunities.
AN
Annelle
1 year ago
I'm cracking up at the typo in Option D. 'sting' instead of 'string'? Somebody needs to proofread these exam questions!
WI
William
1 year ago
User 3: Definitely, that typo could confuse someone during the exam.
KA
Kati
1 year ago
User 2: I know right, they really need to proofread these questions better.
MO
Monroe
1 year ago
User 1: Option D is hilarious, 'sting' instead of 'string'!
AL
Alana
1 year ago
Option C looks promising, but I'm not sure if document.write() is the best way to display the value. Might be better to use innerHTML or textContent.
RA
Rashad
1 year ago
Option C looks promising, but I'm not sure if document.write() is the best way to display the value. Might be better to use innerHTML or textContent.
MA
Marlon
1 year ago
B) Line 2 should be changed to name .prototype . toUpperCase () ;
LY
Lynette
1 year ago
A) Line 2 should be changed to name = name . toUpperCase () ;
ME
Meghan
2 years ago
Haha, Option B is clearly a joke. Why would we need to call toUpperCase() on the prototype? That's just silly.
SE
Selma
1 year ago
Kimberlie: I think so too, calling toUpperCase() directly on the string variable makes more sense.
YE
Yesenia
1 year ago
User 3: Option A seems like the correct answer to me.
KI
Kimberlie
1 year ago
User 2: Yeah, it's definitely not the right choice.
TA
Tasia
1 year ago
User 1: I agree, Option B doesn't make any sense.
AL
Alecia
2 years ago
I think Option D is the way to go. We need to create a new String object to ensure the name is properly defined.
SE
Sena
1 year ago
User 2: Line 1 should be changed to var name = new String(\'Jaccb\').
MI
Micah
1 year ago
User 1: I think Option D is the way to go. We need to create a new String object to ensure the name is properly defined.
TA
Tanja
2 years ago
Option A is the correct answer. We need to call the toUpperCase() method on the name variable to convert it to uppercase.
DA
Daisy
2 years ago
I'm not sure, but I think the answer might be C.
LY
Lyndia
2 years ago
I disagree, I believe the correct answer is B.
MA
Marsha
2 years ago
I think the answer is A.