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NEW QUESTION 34
Given the code fragment:
And given the requirements:
1. Process all the elements of the array in the order of entry.
2. Process all the elements of the array in the reverse order of entry.
3. Process alternating elements of the array in the order of entry.
Which two statements are true? (Choose two.)
Answer: B,C
NEW QUESTION 35
Given the fragment:
String[][] arra = new String[3][];
arra[0] = new String[]{"rose", "lily"};
arra[1] = new String[]{"apple", "berry","cherry","grapes"};
arra[0] = new String[]{"beans", "carrot","potato"};
// insert code fragment here
Which code fragment when inserted at line '// insert code fragment here', enables the code to successfully change arra elements to uppercase?
for (String x:a[]) {B. toUpperCase();
}
}C. for (int i:arra.length) {
for (String x:arra) {
arra[i].toUpperCase();
}
}D. for (int i = 0; i < 3; i++) {
for (int j=0; j < 4; j++) {
arra[i][j] = arra[i][j].toUpperCase();
}
}E. String[][] arra = new String[3][];
arra[0] = new String[]{"rose", "lily"};
arra[1] = new String[]{"apple", "berry","cherry","grapes"};
arra[0] = new String[]{"beans", "carrot","potato"};
for (int i = 0; i < arra.length; i++) {
for (int j=0; j < arra[i].length; j++) {
arra[i][j] = arra[i][j].toUpperCase();
}
}
Answer: A
Explanation:
Incorrect:
not A: arra.length is 3, but the subarrays have 2, 3 and 4 elements. Index will be out of
bound.
not B: The subarrys are of different lengths. Index will be out of bound.
not D: Compile error.
NEW QUESTION 36
Given:
Which approach ensures that the class can be compiled and run?
Answer: B
Explanation:
We need to catch the exception in the doSomethingElse() method.
Such as:
private static void doSomeThingElse() {
try {
throw new Exception();}
catch (Exception e)
{}
}
Note: One alternative, but not an option here, is the declare the exception in
doSomeThingElse and catch it in the doSomeThing method.
NEW QUESTION 37
......
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