CODEWITHSUNDEEP
javaexception
Gianluca Benucci
Gianluca Benucci

Reputation: 135

NumberFormatException - with String of an Hex Value

I got a stream of byte this stream is a bytearray ( byte[] stream ) now i want parse this byte first to an hex value, e.g. 80 = "0x50" then transform this to an int

But for some values like 0xB8 i got an `java.lang.NumberFormatException: For input string: "0xB8"

How could i manage this exception? Maybe with an other type of parse or different type data? 

byte k = stream[i];

        int b = 0;
        try {
            b = Integer.parseInt(String.format("0x%02X",k),16);
        } catch (NumberFormatException e) {
            e.printStackTrace();
        }

Upvotes: 0

Views: 541

Answers (3)

Tix
Tix

Reputation: 614

You must use overloaded parseInt method with radix. For hex value an example: 

Integer.parseInt("84B",16);

Upvotes: 3

Gianluca Benucci
Gianluca Benucci

Reputation: 135

I've resolved my problem with this question

byte k = stream[i];

        int b = 0;
        try {
            String hexStringNumber = String.format("0x%02X",k);
            b = Integer.decode(hexStringNumber);
        } catch (NumberFormatException e) {
            e.printStackTrace();
        }

Upvotes: 0

Nick
Nick

Reputation: 842

Try this:

Integer.parseInt(String.format("0b%02X",k),16);

Binary values are written using "0bxx" syntax

Upvotes: 0

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