CODEWITHSUNDEEP
javaparsinghex
user2303016
user2303016

Reputation: 31

Hexadecimal Conversion in Java Error

I am trying to take in input from users as a string, divide this string into two halfs and convert those strings to hexadecimal ints. I need to use these hexadecimal ints in a TEA encryption algorithm. I am making this code in netbeans for the gui builder.

My Current Code

//Getting initial text
String arg = input.getText();

//establishing Left and Right for TEA
int[] v = new int[2];

// Splitting the string into two.
StringBuilder output2;
for (int x = 0; x < 2; x++ ) {
    output2 = new StringBuilder();
    for (int i = 0; i < arg.length()/2; i++) {
        if (x == 1) 
            output2.append(String.valueOf(arg.charAt(arg.length()/2 + i)));
        else 
            output2.append(String.valueOf(arg.charAt(i)));  
    }
    //converting a half into a string
    String test = output2.toString();
    //printing the string out for accuracy
    System.out.println(test);
    //converting the string to string hex
    test = toHex(test);
    //converting the string hex to int hex.
    v[x] = Integer.parseInt(test, 16);
}


public static String toHex(String arg) {
    return String.format("%x", new BigInteger(arg.getBytes()));
}

I get this error:

java.lang.NumberFormatException: For input string: "6a54657874"

Ive looked around the web for this issue but the error says it is happening when im converting the string to v[x] which multiple sites say is the correct way to put a hex string into an int so i am confused. Please help.

Upvotes: 0

Views: 232

Answers (2)

bdkosher
bdkosher

Reputation: 5883

32 bits is too small for your number. You'll need to use Long.parseLong instead.

Upvotes: 1

jlordo
jlordo

Reputation: 37813

6a54657874 as hex is 456682469492 in decimal. This is bigger than Integer.MAX_VALUE. It will fit into a long.

Make v a long[] and use Long.parseLong(test, 16);

Upvotes: 2

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