Question

analyze the following code:
public class test {
public static void main(string args) {
int n = 10000 10000 10000;
system.out.println(
is \ + n);
}
}

a the result of 10000 10000 10000 is too large to be stored in an int variable n. this causes an underflow and the program is aborted.
b the result of 10000 10000 10000 is too large to be stored in an int variable n. this causes an underflow and the program continues to execute because java does not report errors on underflow.
c the program displays n is 1000000000000.
d the result of 10000 10000 10000 is too large to be stored in an int variable n. this causes an overflow and the program is aborted.
e the result of 10000 10000 10000 is too large to be stored in an int variable n. this causes an overflow and the program continues to execute because java does not report errors on overflow.

Explanation:

1. First, calculate \(10000\times10000\times10000 = 10^{4}\times10^{4}\times10^{4}=10^{12}\). The range of a 32 - bit int in Java is \(-2^{31}\) to \(2^{31}- 1\) (approximately \(-2.147483648\times10^{9}\) to \(2.147483647\times10^{9}\)). Since \(10^{12}\) is larger than the maximum value of int, an overflow occurs.
2. In Java, when an integer overflow happens, the program does not abort. Instead, it continues to execute with the incorrect (wrapped - around) value.
3. Option A and B are wrong because the situation is overflow (not underflow, underflow is for very small values, usually for floating - point numbers or when the number is smaller than the minimum representable value of the type). Option C is wrong because the value will be overflowed and not equal to \(1000000000000\) when stored in an int. Option D is wrong because Java does not abort the program on integer overflow. Option E correctly states that it is an overflow and the program continues to execute as Java does not report errors on integer overflow.

Answer:

E. The result of \(10000 * 10000 * 10000\) is too large to be stored in an int variable n. This causes an overflow and the program continues to execute because Java does not report errors on overflow.