Your Question:

What is the trace of this code after running it?

Step By Step Answers with Explanation

String str = "HELLO-WORLD!!";

String[] a = {"3", "4", "35", "0", "27", "0", "0", "90", "40", "55", "68"};

int x = Integer.parseInt(a[i + 1]);

int y = Integer.parseInt(b[i]);

int x = Integer.parseInt(b[1 - 2]);

int y = Integer.parseInt(a[i]);

} catch (ArrayIndexOutOfBoundsException e) {

System.out.println("Array Out of Bounds");

}

}

1. `String str = "HELLO-WORLD!!";`: This line initializes a string variable `str` with the value "HELLO-WORLD!!".

2. `String[] a = {...};` and `String[] b = {...};`: These lines create two arrays `a` and `b` with some string values.

- `int x = Integer.parseInt(a[i + 1]);`: This line tries to parse `a[i + 1]` as an integer.

- `int y = Integer.parseInt(b[i]);`: This line tries to parse `b[i]` as an integer.

- `int x = Integer.parseInt(b[1 - 2]);`: This line tries to parse `b[-1]`, which will result in an exception.

- `int y = Integer.parseInt(a[i]);`: This line tries to parse `a[i]` as an integer.

- `StringIndexOutOfBoundsException`: This block catches string index out-of-bounds exceptions and prints "String Out of Bounds" if one occurs.

- `NumberFormatException`: This block catches number format exceptions and prints "Can't convert." if one occurs.

4. Parsing `a[1]` as an integer (`x = 4`) and `b[0]` as an integer (`y = 12`) are successful.

5. `System.out.println(x / y);` prints `0` because `4 / 12 = 0`.

- `i = 1`: The code encounters an arithmetic exception when trying to divide by zero (`4 / 0`).

- `i = 2`: The code successfully calculates `35 / 0` and prints `Infinity` (since dividing by zero in Java results in positive or negative infinity).

- `i = 7`: The code encounters an array index out-of-bounds exception when trying to access `b[1 - 2]` (`b[-1]` is out of bounds).

- `i = 8`: The code successfully calculates `90 / 40` and prints `2`.