# What is the result?

Given the code fragment:

What is the result?
A. A B C
B. A B C D E
C. A B D E
D. Compilation fails.

## 2 thoughts on “What is the result?”

1. M says:

The 2D-String array looks like this:
A B C
D E

The outer loop runs and triggers the inner loop. i is currently 0. The inner loop runs and will run for length of arr[0], so 3 times. It prints arr[0][0] + space so “A “ and array[0][1] + space so “B “, and then breaks because of the break statement. Then the outer loop runs again, but now i = 1, so it will run for length or arr[1], which is two times. It will print arr[1][0] + space so “D “ and arr[1][1] + space so “E “. In short, it prints all the elements of the 2D-array in order, but not C because the break-statement prevents the inner loop from printing elements of the first row after it reaches B.

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2. Hack says:

C
public static void main(String[] args) {
String[][] StringArr = {{“A”,”B”,”C”},{“D”,”E”}};
for(int i=0 ; i<StringArr.length;i++){
for(int j=0;j<StringArr[i].length;j++){
System.out.println(StringArr[i][j]+" ");
if(StringArr[i][j].equals("B")){
break; //break the inner loop and go to other iteration i =1
}
}
continue;// is unnecessary as the last statement in a loop
}
}
so A B D E