The line, char newvar = str[4] is declaring a new variable which is of the type character. Further more, when it declares it, it is saying that it is the same as the character at the 4th location in the string above.

The value 4 actually represents the 5th value, since indices start at 0.

Therefore, at str[4] we have the character 'o'.

String s = "abcdefghijklmnop";

String s2 = "";

for(int i = 0; i < s.length(); i += 2) {

     s2 += s.charAt(i+1);

     s2 += s.charAt(i);

}

The key logic of this, of course, occurs in the loop. This loop can be thought of as looking at each pair of characters (hence the use of i += 2 as your increment condition).  Thus, i will be 0,2,4, . . . For 0, you will first place character 1 on s2 then character 0. Then, for 2, you will place 3 followed by 2. The pattern will continue. This is the same as saying that you will first do b, then a, next d, then c, and so forth. Therefore, for this code, you are flipping each pair of letters, thus giving you the answer badcfehgjilknmpo.

for(int i = 0; i < s.length(); i++) {

     if(i != 0) {

          s2 += " ";

     }

     s2 += s;

}

Let us focus on the main for loop that is used in this code. Notice that the loop will run from 0 to the length of the string s. This string is 6 long. Therefore, the loop will run 6 times. The if statement adds a space to the string on every iteration except for the first. Therefore, the code copies the string 6 times, each copy being separated by a space.

The method String.split() splits a String into an array of Strings separated according to the expression within the method argument. The expression String.split("\\s+") splits the String at every whitespace character. The "for" loop concatenates the elements of the String array together, separated by a comma. 

When evaluating a string expression with integers, arithmatic addition is performed if the integers are not separated by string Objects. Since all of the integers and operations in this expression are not separated by String objects, arithmatic operations are performed on the integers and the final airthmatic result is printed to the console.

The presence of the "foo" String in the middle of the expression makes it such that arithmatic operations are performed separately on the left and right sides of the String. It causes each plus sign to perform concatenation instead of addition between terms.

The answer is E. E will throw cause Java to throw an error.

The question tests the students' syntactical knoweldge of Java, as well as the meaning of Java throwing a compiler error. Before running a piece of Java code, it must be compiled. The compiler converts the code into a language the machine understands, but more importantly for this example it catches any syntactical errors that the user might have made. 

Answer A is correct, as you need the new keyword to initialize a String as an Object, so the line of code will throw an error. 

Answer B is incorrect, as Strings can also be initialized as a primitive type. 

Answer C is incorrect, as it is possible to give s a null reference.

Answer D is incorrect, as Java automatically casts the integer 5 to a string, effectivelly performing the following operation: "Hello" + "5". This is called implicit casting. This is a feature of the Java language, but for example, not of Python.

Answer E is incorrect, as Strings are also Objects in Java and can be initialized with the new keyword.

You cannot cast an Object of type Integer to String. The tricky part here is that the compiler will not be able to catch this because it is integerObject has type Object. Had it been of type Integer, a compiler error would be thrown. Thus, the code fragment is allowed to run but an exception is thrown at runtime instead.

Additionally, the option "two" does not make any sense, because even if 2 was cast to a String, the output would be "2", and not "two." 

Also, the option ""2"" does not make sense because the quotes are not displayed.