# AP Computer Science A : Evaluating String Expressions

## Example Questions

### Example Question #31 : Primitive Data Types

string str = "Hello"

char newvar = str;

What is the value of newvar?

'e'

'l'

'H'

" "

'o'

'o'

Explanation:

The line, char newvar = str 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 we have the character 'o'.

### Example Question #41 : Standard Data Structures

String s = "abcdefghijklmnop";

String s2 = "";

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

     s2 += s.charAt(i+1);

     s2 += s.charAt(i);

}

What is the value of s2 at the end of the code above?

adfgbnhciempjkloa

adfgmbnciephjkloa

ponmlkjihgfedcba

badcfehgjilknmpo

aabbccddeeffgghhiijjkkllmmnnoopp

badcfehgjilknmpo

Explanation:

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.

### Example Question #91 : Computer Science

Consider the code below:

String s = "Logic!";

String s2 = "";

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

     if(i != 0) {

          s2 += " ";

     }

     s2 += s;

}

What is the value of s2 at the end of the code's execution?

Logic! Logic! Logic! Logic! Logic! Logic!

Logic! Logic! Logic! Logic! Logic! Logic!

L L L L L L o o o o o o g g g g g g i i i i i i c c c c c c ! ! ! ! ! !

Logic! Logic! Logic!

LoGiC! LoGiC! LoGiC! LoGiC! LoGiC! LoGiC!

Logic! Logic! Logic! Logic! Logic! Logic!

Explanation:

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.

### Example Question #41 : Primitive Data Types

Consider the following code: What is the output of the method call mystery("Green eggs and ham")?

G,r,e,e,n,e,g,g,s,a,n,d,h,a,m,

Greeneggsandham,

ham,

Green,eggs,and,ham,

Green eggs and ham

Green,eggs,and,ham,

Explanation:

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.

### Example Question #42 : Primitive Data Types

What is the output of the following code?

System.out.println( (1+2) + (3-1) + new Integer(5));

36

10

325

45

Syntax error, will not run

10

Explanation:

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.

### Example Question #43 : Primitive Data Types

What is the output of the following code?

System.out.println((2*2) + " foo " + (3-1) + new Integer(5) + "3-2");

4 foo 251

4 foo 253-2

12 foo

4 foo 8

4 foo 73-2

4 foo 253-2

Explanation:

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.

### Example Question #41 : Standard Data Structures

Which of the following will cause the Java compiler to throw an error?

String s = new String("Hello");

String s = "Hello" + 5;

String s = String("Hello");

String s = null;

String s = "Hello";

String s = String("Hello");

Explanation:

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.

### Example Question #1 : Evaluating String Expressions

Suppose you have the following lines of code:

Object integerObj = new Integer(2);

System.out.println((String) integerObj);

What will be the output resulting from running the lines of code?

No output. An exception is thrown of type ClassCastException

two

A compiler error will be thrown

2

"2"

No output. An exception is thrown of type ClassCastException

Explanation:

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.

### All AP Computer Science A Resources 