Why Java Integer Autoboxing Returns the Same Object for Values 0‑127 and How to Compare Correctly

Preface – After Java 5 introduced autoboxing and unboxing, many developers encounter a puzzling behavior where comparing two Integer objects with == yields true for small values but false for larger ones. The article records the issue to avoid future mistakes.

Problem Description

Example 1

public static void main(String[] args) {
    for (int i = 0; i < 150; i++) {
        Integer a = i;
        Integer b = i;
        System.out.println(i + " " + (a == b));
    }
}

Running this prints true for i = 0‑127 and false from 128 onward.

Example 2

public static void main(String[] args) {
    Map
mapA = new HashMap<>();
    Map
mapB = new HashMap<>();
    for (int i = 0; i < 150; i++) {
        mapA.put(i, i);
        mapB.put(i, i);
    }
    for (int i = 0; i < 150; i++) {
        System.out.println(i + " " + (mapA.get(i) == mapB.get(i)));
    }
}

It shows the same true/false pattern.

Analysis – Autoboxing

Autoboxing converts a primitive int to an Integer by calling Integer.valueOf(i) . For values between -128 and 127, valueOf returns a cached object instead of creating a new one.

public static Integer valueOf(int i) {
    if (i >= IntegerCache.low && i <= IntegerCache.high)
        return IntegerCache.cache[i + (-IntegerCache.low)];
    return new Integer(i);
}

The cache is populated in the static block of the inner class IntegerCache :

private static final class IntegerCache {
    static final int low = -128;
    static final int high;
    static final Integer cache[];
    static {
        int h = 127;
        // optional property handling omitted for brevity
        high = h;
        cache = new Integer[(high - low) + 1];
        int j = low;
        for (int k = 0; k < cache.length; k++)
            cache[k] = new Integer(j++);
    }
}

Thus, for -128 ≤ i ≤ 127 the same cached Integer instance is returned, making a == b true. Outside this range a new object is created each time, so a == b becomes false.

Solution

Since == checks object identity, use equals() to compare the numeric values of Integer objects:

for (int i = 0; i < 150; i++) {
    Integer a = i;
    Integer b = i;
    System.out.println(i + " " + a.equals(b));
}

This prints true for all values.

Note

All eight primitive types in Java have corresponding wrapper classes that also use caching for certain ranges (e.g., Byte , Short , Long , Character ), which can lead to similar identity‑comparison surprises.