Converting Integer To Hex Java

Question

I need for my app this int[] format as I wrote in the title.

I'm fetching the value out of a HashMap which is filled out of an XML file with this format [00, 00, 00, 00], but I need it for every int with 0xin front.

I tried to "convert" it via a for-loop:

for(int i = 0; i < value.length; i++){
        value[i] = "0x"  + value[i];
    }

but I cant convert from String to int this way.

Then I tried to change it directly in the XML file, but then my app crashes with an NPE.

Now I want to know if there is a solution to my problem.

EDIT:

For further explanation:

I tried this before, but it did not work:

public int[] getValue(Map map, String key) {
    Map keyMap = map;
    int[] value = {};

    @SuppressWarnings("rawtypes")
    Iterator iter = keyMap.entrySet().iterator();

    while (iter.hasNext()) {
        @SuppressWarnings("rawtypes")
        Map.Entry mEntry = (Map.Entry) iter.next();
        if (mEntry.getKey().equals(key)) {
            value = (int[]) mEntry.getValue();
        }
    }
    return value;
}

Then I tried one of the answers and added

for(int i = 0; i < value.length; i++){
        value[i] = Integer.valueOf(String.valueOf(value[i]), 16);
    }

before the return. Now it works... but I dont know why it works :(

Have you tried searching the internet for "java converting integer to string"? — Smutje
– Smutje, Commented Feb 17, 2014 at 10:40
I dont want a string, I want it like this: int[] int = {0x00, 0x00, 0x00, 0x00} — Fraggles
– Fraggles, Commented Feb 17, 2014 at 10:43
That's not possible, because int is int - hexadecimal prefixes are only for interpretation and not for calculation. — Smutje
– Smutje, Commented Feb 17, 2014 at 10:44
There is no difference between int[] a = {0x00, 0x00, 0x00, 0x00} and int[] a = {0,0,0,0} — Henry
– Henry, Commented Feb 17, 2014 at 10:47
I doubt that value[i] = Integer.valueOf(String.valueOf(value[i]), 16); works in general. The problem is that you already have a wrong value when you read the XML, sometimes even a NumberFormatException. You cannot repair that after the fact. — Henry
– Henry, Commented Feb 17, 2014 at 16:57

Pablo Lozano · Accepted Answer · 2014-02-17 11:32:35Z

2

In any typed language int/number values have no format. The way Java stores int values is as 4 byte number using Two's complement, so 0 or 0x0 or 00 (octal) or 0b is the same value: 32 zeros in a binary word.

SECOND EDIT:

I think you are over-engineering. What you should use is a

Map<String,int[]> map;

then, if you write map.get(key), you'll obtain what you're looking for.

NEW EXPLANATION: The previous solution worked because when you read the XML stored the values as decimal ones:

String s="80";
int i= Integer.valueOf(s); //it stores 80 (decimal value)
int value= Integer.valueOf(String.valueOf(i),16); //It stores 128, or 0x80 (hex value)

edited Feb 17, 2014 at 11:32

answered Feb 17, 2014 at 10:48

Pablo Lozano

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7 Comments

Fraggles Over a year ago

the key is just a String to better identify the value stored to it (for better reading purposes)

Fraggles Over a year ago

okay, the String.valueOf() was part of an answer which is now deleted (dont know why he deleted it). hm... as I now tried to delete the String.valueOf part, it does say The method valueOf(int) in the type Integer is not applicable for the arguments (int, int)

Pablo Lozano Over a year ago

Yes, I was messing myself... let me try to understand again what you need :P

Fraggles Over a year ago

The problem I had was that converted a String out of an XML. String is 80 0 0 0 0 0 0 0. So through a XmlPullparser I fetched the data and stored it in a HashMap. Now I wanted to take via a key the value stored to it. So I fetched it and saved it to the value int array. The Problem why it did not work was, that in my pullparser I didnt convert it correctly to an hex int. But at the moment it works with the answer that was posted, but which is now deleted

Pablo Lozano Over a year ago

@Fraggles Correct, the parsing must be done when reading the XML, using Integer.valueOf(mystring,16) to store the correct integer. You were storing decimal values, that's why you needed to re-evaluate them

|

Triode · Accepted Answer · 2014-02-17 10:39:10Z

1

for(int i = 0; i < value.length; i++){
    value[i] = Integer.valueOf(String.valueOf(value[i]), 16);
}

Try this. If you want the hex value. Or in case if you want hex string you can use below code.

Integer.toHexString(integerValue)

answered Feb 17, 2014 at 10:39

Triode

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2 Comments

Triode Over a year ago

did not get that can you explain further ?

Pablo Lozano Over a year ago

See my answer for more details

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