Java 17 1Z0-829: Core Value Types

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Sample questions

These questions are original IT Mastery practice items aligned to this topic area. They are designed for self-assessment and are not official exam questions.

Question 1

Topic: Handling Date, Time, Text, Numeric, and Boolean Values

A developer is checking how a small mixed numeric expression behaves. What is printed by this Java 17 program?

class Demo {
    public static void main(String[] args) {
        byte b = 4;
        short s = 3;
        char c = 2;
        long l = 4L;
        float f = 1.0f;
        boolean result = (b += s) > c + l && f + l == 5.0f;
        System.out.println(result + ":" + b);
    }
}

Options:

• A. true:7

• B. false:7

• C. true:4

• D. The program does not compile.

Best answer: A

Explanation: Compound assignment performs the arithmetic using numeric promotion and then casts the result back to the left-hand type. Here, b += s stores 7 in b. The later mixed numeric comparisons are both true, so the program prints true:7.

The core rule is binary numeric promotion plus compound assignment conversion. A compound assignment such as b += s is roughly equivalent to b = (byte)(b + s), so it compiles even though b + s alone would have type int. After this assignment, b is 7. Then c + l is promoted to long, giving 6L, so 7 > 6L is true. Also, f + l is promoted to float, giving 5.0f, which equals 5.0f. Since both sides of && are true, result is true and the updated b value is printed.

• Compilation error trap fails because compound assignment can implicitly cast the arithmetic result back to the left-hand type.
• False comparison trap fails because 7 > 6L and 5.0f == 5.0f both evaluate to true.
• Original value trap fails because b += s mutates b before System.out.println() runs.

Question 2

Topic: Handling Date, Time, Text, Numeric, and Boolean Values

A team is refactoring this Java 17 method. It should reverse a non-empty code, add a leading #, remove the last character after the reversal, and append !. For input ABCD, the expected result is #DCB!, but the current method fails at runtime. Which change is the best fix?

static String tag(String code) {
    var sb = new StringBuilder(32);
    sb.append(code)
      .reverse()
      .insert(0, '#')
      .delete(sb.capacity() - 1, sb.capacity())
      .append('!');
    return sb.toString();
}

Options:

• A. Replace StringBuilder with StringBuffer and keep the chain unchanged.

• B. Replace the delete call with .deleteCharAt(sb.capacity() - 1).

• C. Replace the delete call with .delete(sb.length() - 1, sb.length()).

• D. Initialize sb with new StringBuilder(code) and remove .append(code).

Best answer: C

Explanation: StringBuilder operations such as append, reverse, insert, delete, and deleteCharAt mutate the same builder and return it for chaining. Character operations use valid indexes based on length(), not capacity(). The capacity is only storage space and can be larger than the number of characters.

The defect is using capacity() as if it were the last valid character index. After appending ABCD, reversing, and inserting #, the builder content is #DCBA and its length() is 5. Its capacity is still much larger, so attempting to delete near capacity() - 1 uses an invalid character index and fails. delete(start, end) removes characters from start up to but not including end, so delete(sb.length() - 1, sb.length()) removes only the current last character. The chained calls are fine because these methods mutate and return the same StringBuilder.

• Constructor capacity fails because new StringBuilder(code) still gives extra capacity and does not make capacity equal to the current content length.
• StringBuffer swap solves synchronization, not the invalid index caused by using capacity.
• deleteCharAt with capacity still uses an invalid character index because capacity is not the number of stored characters.

Question 3

Topic: Handling Date, Time, Text, Numeric, and Boolean Values

A service compares a cached status code with a value read from a request. Status text should match when the characters are the same, regardless of whether a String came from a literal or was constructed at runtime. Two null status values should be treated as matching, and numeric retry counts should be compared by value.

import java.util.Objects;

class StatusCheck {
    static boolean matches(String cached, String received,
                           int attempts, int limit) {
        return cached == received && attempts == limit;
    }
}

Which refactor is the best correction?

Options:

• A. Use return cached == received && Integer.valueOf(attempts) == Integer.valueOf(limit);

• B. Use return Objects.equals(cached, received) && attempts == limit;

• C. Use return cached.intern() == received.intern() && attempts == limit;

• D. Use return cached.equals(received) && Objects.equals(attempts, limit);

Best answer: B

Explanation: The defect is using == for String values, which checks reference identity rather than character equality. Objects.equals(cached, received) gives null-safe String equality, and primitive int values should continue using == for numeric comparison.

For objects, == compares references; for primitives, == compares values. String literals are interned, so two identical literals may share one reference and make == seem to work, but a String built at runtime can have the same characters and a different reference. Objects.equals(cached, received) safely handles both normal String.equals() comparison and nulls. The retry counts are primitive int values, so attempts == limit already compares their numeric values directly. The key refactor changes only the object equality check and leaves the primitive equality check alone.

• Interning workaround fails because calling intern() on null throws and interning is not the right general-purpose equality test.
• Direct equals call compares characters but can throw when cached is null, and boxing the int values is unnecessary.
• Wrapper references leave the String reference comparison defect and add another reference-equality comparison for boxed integers.

Question 4

Topic: Handling Date, Time, Text, Numeric, and Boolean Values

During troubleshooting, a developer runs this Java 17 diagnostic and gets true false true:

public class Probe {
    public static void main(String[] args) {
        int p = 7, q = 7;
        String a = "OK";
        String b = new String("OK");
        String c = b.intern();

        System.out.println((p == q) + " " + (a == b) + " " + (a == c));
    }
}

What is the best explanation for the unexpected middle value?

Options:

• A. a == b compares two different object references.

• B. intern() creates a new nonliteral String object.

• C. Primitive int values are compared by reference.

• D. String values use equals() automatically with ==.

Best answer: A

Explanation: The middle comparison is false because == on String compares references, not character content. The literal "OK" is interned, while new String("OK") creates a separate object. Use equals() when the goal is string content comparison.

Java uses == differently for primitives and references. For primitives such as int, == compares numeric values, so p == q is true. For object references such as String, == compares whether both variables refer to the same object. The literal "OK" refers to the interned string, but new String("OK") creates a separate String object with the same contents, so a == b is false. Calling b.intern() returns the canonical interned instance, so a == c is true. The key takeaway is to use a.equals(b) for string content comparison.

• Primitive confusion fails because primitive int equality compares values, not references.
• Interning reversal fails because intern() returns the canonical pooled string rather than a fresh nonliteral object.
• Automatic equals fails because == does not call String.equals() implicitly.

Question 5

Topic: Handling Date, Time, Text, Numeric, and Boolean Values

No imports are needed. Given this Java 17 code, what is the result?

public class Limits {
    public static void main(String[] args) {
        int min = -2_147_483_648;
        long next = 2_147_483_648;
        long fixed = 2_147_483_648L;
        System.out.println(min + "," + next + "," + fixed);
    }
}

Options:

• A. It prints -2147483648,2147483648,2147483648.

• B. Compilation fails; 2_147_483_648 needs an L suffix.

• C. It throws ArithmeticException before printing.

• D. Compilation fails at int min; the value is below int range.

Best answer: B

Explanation: Integer literals without a suffix are treated as int literals first. The positive literal 2_147_483_648 is not valid in that long assignment unless it has an L suffix, even though the variable type is long.

Java determines a numeric literal’s type before considering the assignment target. An unsuffixed decimal integer literal is an int literal, with a special allowance for -2_147_483_648 so Integer.MIN_VALUE can be written directly. However, the positive literal 2_147_483_648 cannot be assigned to long unless it is written as a long literal, such as 2_147_483_648L.

The fixed declaration is valid, but compilation stops because the earlier next declaration is invalid.

• Printed values assumes the long target type changes the literal type, but literal validity is checked first.
• Minimum int is valid here because Java permits -2_147_483_648 as the int minimum value.
• Runtime exception confuses compile-time literal checking with runtime arithmetic behavior.

Question 6

Topic: Handling Date, Time, Text, Numeric, and Boolean Values

A scheduling test fails on the U.S. DST fallback. The next run must be the same New York wall-clock time on the next calendar day. This code prints 2024-11-04T00:30-05:00[America/New_York], but the expected value is 2024-11-04T01:30-05:00[America/New_York].

var zone = ZoneId.of("America/New_York");
var start = ZonedDateTime.ofLocal(
    LocalDateTime.of(2024, 11, 3, 1, 30),
    zone,
    ZoneOffset.of("-04:00"));
var next = start.toInstant()
                .plus(Duration.ofDays(1))
                .atZone(zone);
System.out.println(next);

What is the best fix?

Options:

• A. Store the time with fixed offset -04:00.

• B. Use start.plusDays(1) and convert afterward.

• C. Use start.toInstant().plus(Period.ofDays(1)).

• D. Call withZoneSameInstant(zone) before adding.

Best answer: B

Explanation: The bug comes from adding a 24-hour duration to an Instant, which is a timeline calculation. The requirement is a local calendar-day adjustment in America/New_York, so the addition should be done on the ZonedDateTime before converting to an Instant.

Instant represents a point on the UTC timeline, so Duration.ofDays(1) means exactly 24 hours later. Starting at 2024-11-03T01:30-04:00, adding 24 hours lands at 2024-11-04T05:30Z, which is 00:30 in New York after the offset changes to -05:00. For “same local time tomorrow,” add the calendar day in the zone: start.plusDays(1). That adjusts the local date-time to November 4 at 01:30, then resolves the correct zone offset. The key distinction is timeline duration versus zone-aware calendar adjustment.

• Period on Instant still adjusts an instant, not a New York wall-clock calendar date.
• Fixed offset ignores the zone rule change to -05:00 after the fallback.
• Same instant zone call does not change the value because the ZonedDateTime already uses that zone.

Question 7

Topic: Handling Date, Time, Text, Numeric, and Boolean Values

No imports are needed. What is the result of compiling and running the following Java 17 code?

public class TextTest {
    static void normalize(String s) {
        s.trim();
        s.replace("java", "Java");
        s += "!";
    }

    public static void main(String[] args) {
        String msg = " java ";
        normalize(msg);
        System.out.print("[" + msg + "]");
    }
}

Options:

• A. It does not compile.

• B. It prints [Java!].

• C. It prints [ java ].

• D. It prints [ java !].

Best answer: C

Explanation: String objects are immutable in Java. Methods such as trim() and replace() return new String values rather than modifying the existing object, and this code ignores those return values.

The core rule is String immutability. The parameter s initially refers to the same String object as msg, but calling s.trim() and s.replace(...) does not change that object. Each method returns a new String when a change is needed, and the returned values are discarded here. The statement s += "!" is also a reassignment of the local parameter s to a new String; it does not reassign the caller’s variable msg. After normalize(msg) finishes, msg still refers to the original string containing one leading and one trailing space.

• Mutable String assumption fails because trim() and replace() do not alter the original String object.
• Caller reassignment assumption fails because s += "!" updates only the local parameter variable.
• Compilation failure is not supported because the snippet uses valid Java 17 syntax and no imports are required.

Question 8

Topic: Handling Date, Time, Text, Numeric, and Boolean Values

What is the output of this Java 17 code? Each option lists the printed lines in order.

import java.time.*;
import java.time.temporal.ChronoUnit;

public class Dates {
    public static void main(String[] args) {
        LocalDate start = LocalDate.of(2024, 3, 10);
        start.plusDays(2);
        LocalDate end = start.withMonth(4).plusDays(5);

        System.out.println(start);
        System.out.println(end);
        System.out.println(ChronoUnit.DAYS.between(start, end));
        System.out.println(start.until(end).getDays());
    }
}

Options:

• A. 2024-03-10; 2024-04-15; 5; 5

• B. 2024-03-12; 2024-04-17; 36; 5

• C. 2024-03-10; 2024-04-15; 36; 36

• D. 2024-03-10; 2024-04-15; 36; 5

Best answer: D

Explanation: Java date-time classes such as LocalDate are immutable, so start.plusDays(2) has no effect unless its result is assigned. ChronoUnit.DAYS.between(start, end) returns the total number of days, while start.until(end).getDays() returns only the day component of the resulting Period.

The core rule is immutability plus the difference between a unit-based count and a Period. The ignored plusDays(2) call does not change start, so it remains 2024-03-10. The expression start.withMonth(4).plusDays(5) creates 2024-04-15. From March 10 to April 15, ChronoUnit.DAYS.between counts 36 total days. However, start.until(end) is a Period of 1 month and 5 days, so getDays() returns 5, not 36. The key takeaway is that Period.getDays() is a component value, not a total duration in days.

• Mutation assumption fails because plusDays(2) returns a new LocalDate and does not modify start.
• Total Period days fails because Period.getDays() returns only the day component after months are accounted for.
• Unit-count confusion fails because ChronoUnit.DAYS.between counts total days, not the Period day component.

Question 9

Topic: Handling Date, Time, Text, Numeric, and Boolean Values

Which statement is correct for Java 17 primitive numeric expressions?

Options:

• A. Casting a double to an int rounds to nearest.

• B. Integer division by zero evaluates to Infinity.

• C. Combining int and float promotes the result to double.

• D. Overflow in int addition wraps around silently.

Best answer: D

Explanation: Java integer arithmetic has defined overflow behavior: it silently wraps around for operations such as int addition. This is different from integer division by zero, which throws ArithmeticException, and from floating-point promotion rules.

For primitive integer arithmetic, Java does not signal overflow for operations such as addition, subtraction, or multiplication. If an int addition exceeds the range of int, the result is the low-order 32 bits of the mathematical result, commonly described as wraparound. Separate rules handle other numeric cases: integer division truncates toward zero, integer division by zero throws ArithmeticException, and binary numeric promotion with int and float produces a float result.

The key distinction is that overflow is silent, while integer division by zero is not.

• Division by zero fails because int division by zero throws ArithmeticException, not Infinity.
• Narrowing conversion fails because a double cast to int truncates toward zero, not nearest.
• Promotion rule fails because int combined with float promotes to float, not double.

Question 10

Topic: Handling Date, Time, Text, Numeric, and Boolean Values

A developer compares a Java 17 text block with a regular string literal. What is printed?

public class TextCheck {
  public static void main(String[] args) {
    String a = """
        red
          blue\s
        green\
        """;
    String b = "red\n  blue \ngreen";
    System.out.print(a.equals(b) + ":" + a.length());
  }
}

Options:

• A. It prints false:17.

• B. It prints true:18.

• C. It prints false:18.

• D. It prints true:17.

Best answer: D

Explanation: The text block content matches the regular string literal exactly. Java removes the common incidental indentation, translates \s to one space, and uses the trailing \ to omit the line terminator after green.

Text blocks are processed by normalizing line terminators, removing incidental indentation, and then translating escape sequences. Here, the common leading indentation is removed from each content line, leaving red, two spaces before blue, and green. The \s escape contributes a real space after blue; using an actual trailing space in source would be unreliable because trailing whitespace is stripped. The backslash at the end of the green line suppresses that line terminator, so there is no final newline.

The resulting value is red\n blue \ngreen, whose length is 17 characters.

• Extra newline fails because the trailing \ removes the line terminator before the closing delimiter.
• Missing preserved space fails because \s becomes one actual space after blue.
• Indentation mismatch fails because common incidental indentation is stripped from the text block.

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