Why does this Java parseInt hex string result in NumberFormatException?

Question

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)

@John3136 - The spec doesn't mention the 0x prefix at all. — Polynomial
– Polynomial, Commented Feb 8, 2012 at 9:34
You are probably right - 0x would actually be on an int v=0xabcd rather than in a string. — John3136
– John3136, Commented Feb 8, 2012 at 9:38

David Conrad · Accepted Answer · 2015-02-04 23:30:32Z

6

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.

edited Feb 4, 2015 at 23:30

David Conrad

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answered Feb 8, 2012 at 9:36

Polynomial

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3 Comments

Ignoreme Over a year ago

Thanks, I wasn't really thinking about that. I confused the signed and unsigned int bounds. Unfortunately I'll have to test if parsing to long, and converting back to int effects the values of the bits (pesky sign bit!)

Ignoreme Over a year ago

Surprisingly no, the bits all stay consistent through the whole conversion, process. Useful!

Louis Wasserman Over a year ago

You might find Guava's UnsignedInts.parseUnsignedInt relevant.

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