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Home/ Questions/Q 747399
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T14:12:01+00:00 2026-05-14T14:12:01+00:00

While looking in the code of the method: Integer.toHexString I found the following code

  • 0

While looking in the code of the method:

Integer.toHexString

I found the following code :

public static String toHexString(int i) {
    return toUnsignedString(i, 4);
}

private static String toUnsignedString(int i, int shift) {
    char[] buf = new char[32];
    int charPos = 32;
    int radix = 1 << shift;
    int mask = radix - 1;
    do {
        buf[--charPos] = digits[i & mask];
        i >>>= shift;
    } while (i != 0);

    return new String(buf, charPos, (32 - charPos));
}

The question is, in toUnsignedString, why we create a char arr of 32 chars?

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  1. Editorial Team
    Editorial Team
    2026-05-14T14:12:02+00:00Added an answer on May 14, 2026 at 2:12 pm

    32 characters is how much you need to represent an int in binary (base-2, shift of 1, used by toBinaryString).

    It could be sized exactly, but I guess it has never made business sense to attempt that optimisation.

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