Java Output Prediction Questions 2026: 30 Tricky Snippets

Last Updated: June 2026 | Level: Freshers to Mid-Level | Read Time: ~16 min

Predict-the-output questions are how interviewers test whether you truly understand Java semantics rather than memorising definitions. Candidates report these dominate aptitude-style coding rounds and MCQ screens. This guide gives 30 carefully chosen snippets with the exact output and a precise explanation of why. Every behaviour shown is verified against the official Oracle Java specification; confirm any edge case on the JLS, since output traps often hinge on a single spec rule.

Pair this with Java Interview Questions 2026 and Java String Interview Questions 2026.

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Table of Contents

1. Operators and Precedence (Q1 to Q8)
2. Autoboxing and Caching (Q9 to Q14)
3. Static and Inheritance (Q15 to Q22)
4. Strings and Misc (Q23 to Q30)
5. Quick Trap Table
6. Frequently Asked Questions

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Operators and Precedence

Q1.

int x = 5;
System.out.println(x++ + ++x);

Output: 12 Why: x++ yields 5 (then x becomes 6), ++x makes x 7 and yields 7. 5 + 7 = 12.

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Q2.

System.out.println(10 + 20 + "Java" + 10 + 20);

Output: 30Java1020 Why: Left to right: 10 + 20 = 30, then string concatenation begins, so the trailing numbers append as text.

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Q3.

int a = 1;
System.out.println(a == 1 ? "one" : a == 2 ? "two" : "other");

Output: one Why: The ternary is right-associative but evaluates the first condition first; a == 1 is true, so "one".

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Q4.

System.out.println(5 / 2);
System.out.println(5.0 / 2);
System.out.println(5 % 2);

Output:

2
2.5
1

Why: Integer division truncates to 2. Mixing with a double promotes to 2.5. Modulus gives the remainder 1.

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Q5.

boolean b = false;
if (b = true) System.out.println("yes"); else System.out.println("no");

Output: yes Why: b = true is assignment, not comparison; it assigns true and the expression evaluates to true. A frequent typo-based trap.

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Q6.

int i = 10;
System.out.println(i > 5 & i++ > 8);
System.out.println(i);

Output:

true
11

Why: & is non-short-circuit, so both sides evaluate; i++ runs and i becomes 11.

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Q7.

System.out.println(1 << 3);
System.out.println(-8 >> 1);
System.out.println(-8 >>> 28);

Output:

8
-4
15

Why: 1 << 3 = 8. Signed right shift preserves the sign giving -4. Unsigned right shift fills with zeros, exposing the high bits as 15.

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Q8.

char c = 'A';
System.out.println(c + 1);
System.out.println((char)(c + 1));

Output:

66
B

Why: c + 1 promotes char to int (65 + 1). Casting back to char yields 'B'.

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Autoboxing and Caching

Q9.

Integer a = 100, b = 100;
Integer c = 200, d = 200;
System.out.println(a == b);
System.out.println(c == d);

Output:

true
false

Why: Integer caches -128 to 127, so 100 reuses one object (==true). 200 is outside the cache, so two distinct objects (==false). The single most asked output trap.

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Q10.

Integer x = null;
int y = x;

Output: NullPointerException Why: Unboxing a null Integer to a primitive int calls x.intValue() on null, throwing NPE at runtime.

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Q11.

Long l = 1L;
Integer i = 1;
System.out.println(l.equals(i));

Output: false Why: equals checks type first; a Long is never equal to an Integer even with the same numeric value.

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Q12.

int i = 1000;
Integer boxed = i;
System.out.println(boxed.equals(1000));

Output: true Why: equals compares value, not identity, so it ignores the cache and returns true.

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Q13.

double d = 0.1 + 0.2;
System.out.println(d == 0.3);
System.out.println(d);

Output:

false
0.30000000000000004

Why: Binary floating point cannot represent 0.1 and 0.2 exactly, so the sum has a tiny error. Never compare doubles with ==; use a tolerance.

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Q14.

System.out.println(0.1 + 0.2 == 0.3);
System.out.println(Math.abs(0.1 + 0.2 - 0.3) < 1e-9);

Output:

false
true

Why: Direct equality fails due to representation error; an epsilon comparison is the correct pattern.

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Static and Inheritance

Q15.

class A {
    A() { print(); }
    void print() { System.out.println("A"); }
}
class B extends A {
    int x = 5;
    void print() { System.out.println("x=" + x); }
}
new B();

Output: x=0 Why: The base constructor calls the overridden print before B's field initialiser runs, so x is still 0.

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Q16.

class P { static void who() { System.out.println("P"); } }
class C extends P { static void who() { System.out.println("C"); } }
P ref = new C();
ref.who();

Output: P Why: Static methods are not polymorphic; resolution uses the reference's static type.

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Q17.

class A { int x = 1; }
class B extends A { int x = 2; }
A a = new B();
System.out.println(a.x);

Output: 1 Why: Fields are not polymorphic; field access uses the static type of the reference, so A's x (1) is read.

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Q18.

static int count;
static { count = 5; }
public static void main(String[] a) { System.out.println(count); }

Output: 5 Why: The static initialiser block runs at class load, before main, setting count to 5.

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Q19.

class A {
    static int x = 10;
    static { x = 20; }
}
System.out.println(A.x);

Output: 20 Why: Static field initialisers and static blocks run top to bottom; the block runs after the field assignment, leaving 20.

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Q20.

int test() {
    try { return 1; }
    finally { return 2; }
}

Output: 2 Why: A return in finally overrides the try's return value.

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Q21.

class Base { Base() { System.out.print("B"); } }
class Derived extends Base { Derived() { System.out.print("D"); } }
new Derived();

Output: BD Why: The implicit super() runs the Base constructor first, then the Derived body.

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Q22.

final int[] arr = {1, 2, 3};
arr[0] = 99;
System.out.println(arr[0]);

Output: 99 Why: final makes the reference unchangeable, not the contents. You can still mutate elements.

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Strings and Misc

Q23.

String a = "x";
String b = "x";
System.out.println(a == b);

Output: true Why: Both literals share the same pooled reference.

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Q24.

String s = "abc";
s.toUpperCase();
System.out.println(s);

Output: abc Why: Strings are immutable; the returned uppercase string is discarded.

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Q25.

System.out.println("a,b,,".split(",").length);

Output: 2 Why: Trailing empty strings are dropped by default split, leaving "a" and "b".

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Q26.

List<Integer> list = new ArrayList<>(List.of(1, 2, 3));
list.remove(1);
System.out.println(list);

Output: [1, 3] Why: remove(int) removes by index, so index 1 (value 2) is removed, not the value 1.

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Q27.

System.out.println(Integer.parseInt("10", 2));

Output: 2 Why: The second argument is the radix; "10" in base 2 equals decimal 2.

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Q28.

int x = 5;
switch (x) {
    case 5: System.out.print("five ");
    case 6: System.out.print("six ");
    default: System.out.print("def");
}

Output: five six def Why: No break, so execution falls through every subsequent case.

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Q29.

System.out.println((Object) "1" == (Object) "1");
System.out.println(Integer.valueOf(127) == Integer.valueOf(127));

Output:

true
true

Why: Both string literals are pooled, and 127 is within the Integer cache, so both comparisons share references.

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Q30.

int i = 0;
i = i++;
System.out.println(i);

Output: 0 Why: i++ returns the old value 0, which is then assigned back to i, overwriting the increment. Classic self-assignment trap.

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Quick Trap Table

Trap	Rule to remember
Integer cache	-128 to 127 cached, == true; outside, false
Unbox null	NPE on Integer to int when null
Fields	not polymorphic, use static type
Static methods	hidden, not overridden
finally return	overrides try return
split trailing	empty strings dropped
list.remove(int)	removes by index, not value
i = i++	increment lost
double ==	representation error, use epsilon

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Frequently Asked Questions

Are predict-the-output questions common in Java interviews in 2026?

Yes. Candidates report that aptitude-style coding rounds at service companies and MCQ screens at many product companies lean heavily on output prediction for autoboxing, integer caching, and operator precedence.

What is the most common output trap in Java?

The Integer cache trap: values from -128 to 127 are cached so == returns true, but outside that range each box is a new object so == returns false. This single behaviour catches the most candidates.

How do I get fast at predicting output?

Trace evaluation strictly left to right, watch the static versus runtime type for method dispatch, and memorise the special cases: integer cache, string pool, finally over return, and floating point rounding.

Why does i = i++ leave i unchanged?

i++ evaluates to the old value before incrementing, and that old value is then assigned back to i, overwriting the incremented result. The post-increment side effect is lost to the assignment.

Should I write break in every switch case?

In the classic switch statement, yes, unless you intend fall-through. The newer arrow-form switch expression does not fall through, which removes this entire class of bug.

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Related Articles

• Java Interview Questions 2026, the master 50-question set
• Java String Interview Questions 2026, string-pool traps in depth
• Java OOPs Interview Questions 2026, inheritance and dispatch
• Java Collections Interview Questions 2026, list and map behaviour

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Every output above is verified against the official Oracle Java specification. Confirm any edge case on the JLS. This guide reflects candidate-reported patterns and public preparation resources as of June 2026.