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Quw from KB book, chap 4 -> test Q 3

 
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Can anyone explain me this code?

1. class Comp2 {
2. public static void main(String[] args) {
3. float f1 = 2.3f;
4. float[][] f2 = {{42.Of}, {l.7f, 2.3f}, {2.6f, 2.7f}};
5. float[] f3 = {2.7f};
6. Long x = 42L;
7. // insert code here
8. System.out.println("true");
9. }
10. }

And the following five code fragments:

F1. if (f1 == f2)
F2. if (f1 == f2[2][1])
F3. if (x == f2[0][0])
F4. if (f1 == f2 [1,1] )
F5. if (f3 == f2 [2] )

What is true?

A. One of them will compile, only one will be true.

B. Two of them will compile, only one will be true.

C. Two of them will compile, two will be true.

D. Three of them will compile, only one will be true.

E. Three of them will compile, exactly two will be true.

F. Three of them will compile, exactly three will be true.


the answer is D
but i dont understand the reason given in the answer.
please explain...
thanks
 
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which part you don't understand?


F1. if (f1 == f2)// wont compile because f1 is a float variable and f2 is 2d array of floats
F2. if (f1 == f2[2][1])//will compile comparing f1 value with f2[2][1] results in false
F3. if (x == f2[0][0])// will compile and results in true
F4. if (f1 == f2 [1,1] )// syntax error
F5. if (f3 == f2 [2] )// will compile comparing 2 array objects results in false

so 3 will compile and only one will result in true
[ April 10, 2008: Message edited by: sridhar row ]
 
Sanket Modi
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thnks. i understood.
i was confused with F3...
 
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i didn't get this

F3. if (x == f2[0][0])// will compile and results in true


Please explain
 
Don't get me started about those stupid light bulbs.
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