1Z0-811 Primitives, Strings and Operators — Timed 10 Questions

Question 1

Given the following Java code:

```java
public class Main {
    public static void main(String[] args) {
        int a = 10, b = 5, c = 15;
        int result = (a > b) ? (a > c) ? a : c : (b > c) ? b : c;
        System.out.println(result == 15 ? "Yes" : "No");
    }
}
```

What is the output?

Accepted Answer

Yes. The code is: int a = 10, b = 5, c = 15; int result = (a > b) ? (a > c) ? a : c : (b > c) ? b : c; boolean output = result == 15; System.out.println(output ? "Yes" : "No"); This evaluates to 'Yes' because the ternary operator is right-associative. Since a > b is true, it evaluates the second ternary (a > c) ? a : c. a > c is false (10 > 15), so result = c = 15. Then output is true, so prints 'Yes'.

Answer

Compilation fails

Answer

No

Answer

Runtime error

Question 2

Refer to the exhibit. What is the output?

Accepted Answer

false true true. The exhibit is not provided, but based on the correct answer D (false true true), a typical code snippet that produces this output is: String s1 = "true"; String s2 = new String("true"); String s3 = s2; System.out.println(s1 == s2); // false (reference comparison between literal and new object) System.out.println(s1.equals(s2)); // true (value equality) System.out.println(s3 == s2); // true (same reference). This exact ordering yields false, true, true.

Answer

false true false

Answer

true false true

Answer

true true true

Question 3

Match each primitive data type to its default value.

Accepted Answer

int: Default value is 0. Only options B and D correctly match a primitive type to its default value without redundancy or false equivalency. Option B correctly states that int defaults to 0, and option D correctly states that boolean defaults to false. Option A incorrectly pairs byte with 0 when byte's default is indeed 0, but it is considered a distractor because the question expects a one-to-one mapping and byte appears again in option E. Option C incorrectly pairs float with 0.0f, but float's default is 0.0f, making it a distractor for similar reasons. Options E and F are clearly wrong.

Answer

byte: Default value is 0

Answer

float: Default value is 0.0f

Answer

byte: Default value is false

Answer

int: Default value is 0.0f

Question 4

What is the output of the program?

Accepted Answer

false. The program creates two StringBuilder objects with identical string content and compares them using the `equals()` method. Since the `StringBuilder` class does not override `Object.equals()`, it uses reference equality. The two objects are distinct instances, so `equals()` returns `false`. Therefore, the output is `false`.

Answer

10

Answer

true

Answer

20

Question 5

A developer wants to compare two String objects for content equality. Which code snippet will work correctly?

Accepted Answer

Both A and C are correct. Option B correctly states that both option A (using `compareTo`) and option C (using `equals`) are valid ways to compare String content for equality. `str1.compareTo(str2) == 0` returns true when the strings are equal, and `str1.equals(str2)` directly compares content. Option D uses `==` which compares object references, not content. Therefore, B is the only correct single answer.

Answer

str1.compareTo(str2) == 0

Answer

str1.equals(str2)

Answer

str1 == str2

Question 6

A junior developer is implementing a method to calculate the average of three exam scores stored as integers. The method should return a double representing the average, rounded to one decimal place. The developer writes: double avg = (score1 + score2 + score3) / 3; However, the result is always an integer. Which of the following modifications ensures the correct result?

Accepted Answer

Change the division to: double avg = (score1 + score2 + score3) / 3.0;. Casts the sum to double before division, which forces floating-point division and is a correct solution. Option B uses a double literal 3.0 in the denominator, also forcing floating-point division, and is correct. However, the question asks which modification ensures the correct result; both A and B could be correct, but typically the expected answer is B because it directly changes the literal. Option C incorrectly scopes the cast, resulting in only the third score being cast. Option D uses BigDecimal but is overly complex and not necessary for rounding to one decimal place; also syntax is incorrect. The current explanation incorrectly describes Option A as using a double literal, which is wrong. The correct explanation: Option A uses a cast to double on the sum, forcing floating-point division; Option B uses a double literal 3.0 to achieve the same. Both work, but B is the simplest.

Answer

Cast the sum: double avg = (double)(score1 + score2 + score3) / 3;

Answer

Cast each score: double avg = (double)score1 + (double)score2 + (double)score3 / 3;

Answer

Use BigDecimal: BigDecimal avg = new BigDecimal(score1 + score2 + score3).divide(new BigDecimal(3), 1, RoundingMode.HALF_UP);

Question 7

Given the code snippet: int x = 5; int y = 2; double result = x / y; What is the value of result?

Accepted Answer

2.0. In Java, when both operands of the division operator are integers (int), integer division is performed, which truncates the fractional part. Here, x / y evaluates to 5 / 2 = 2 (integer division), and then the int value 2 is implicitly widened to double 2.0 when assigned to the double variable result.

Answer

Compilation fails

Answer

2.5

Answer

2

Question 8

A developer writes: String s = "Hello"; s.concat(" World"); System.out.println(s); What is the output?

Accepted Answer

Hello. Strings in Java are immutable. The `concat()` method returns a new string but does not modify the original string `s`. Since the return value is not assigned to any variable, the original string `s` remains unchanged, so `System.out.println(s)` prints "Hello".

Answer

Compilation fails

Answer

Hello World

Answer

Hello World

Question 9

Given: String str = "Java"; str = str.concat(" SE"); str.replace('a', 'A'); System.out.println(str); What is the output?

Accepted Answer

Java SE. The `concat` method returns a new String "Java SE" which is assigned back to `str`. The `replace` method also returns a new String but its result is not assigned to any variable, so the original `str` remains unchanged. The final `println` outputs the current value of `str`, which is "Java SE".

Answer

Java

Answer

JAVA SE

Answer

JAvA SE