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Editorial Team
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Asked: 2026-05-29T11:19:44+00:00

Integer.parseInt(ff8ca87c, 16); This gives me a NumberFormatException for some reason. Do you know why

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Integer.parseInt("ff8ca87c", 16);

This gives me a NumberFormatException for some reason. Do you know why that is?

Exception in thread "main" java.lang.NumberFormatException: For input string: "ff8ca87c"
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at java.lang.Integer.valueOf(Unknown Source)
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1 Answer

  1. Editorial Team

    The reason it fails is that you’re trying to put +0xff8ca87c into a signed integer. The maximum value of a 32-bit signed integer is +0x7fffffff, because the most significant bit is used to store the sign.

    Try using a long instead. The maximum value of a 64-bit signed int is 0x7fffffffffffffff, which is more than adequate for your needs in this case.

    Or, in Java 8 you can use Integer.parseUnsignedInt("ff8ca87c", 16); which will treat the value as an unsigned integer.

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