Convert List to string

Question

I have the following code:

List<String> decryptedPasswordInPairs = new ArrayList<String>();
String A5 = "A5";
for (String oddPair : oddPairsInEncryptedPassword) {
    List<Byte> sample = new ArrayList<>();
    for (int i = 0; i < oddPair.length(); i++) {
        byte x = (byte) (oddPair.charAt(i) ^ A5.charAt(i % A5.length()));
        sample.add(x);
    }
    String result = sample.toString();
    decryptedPasswordInPairs.add(result);
}

In this code, result is displayed as [0,7] instead of 07 when I debug the program to check the value of result.

Is there a way to convert this List of Bytes to a String without having any problems?

Ramsay Domloge · Accepted Answer · 2016-07-02T15:28:37.810

2

In this solution, I collect the characters in a StringBuilder and convert to a String at the end:

List<String> decryptedPasswordInPairs = new ArrayList<String>();
    String A5 = "A5";

    List<String> oddPairsInEncryptedPassword = new LinkedList<String>();
    oddPairsInEncryptedPassword.add("A2");

    for (String oddPair : oddPairsInEncryptedPassword) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < oddPair.length(); i++) {
            byte x = (byte) (oddPair.charAt(i) ^ A5.charAt(i % A5.length()));
            sb.append(""+x);
        }

        String result = sb.toString();
        System.out.println(result);
        decryptedPasswordInPairs.add(result);
    }

edited Jul 02 '16 at 15:28

answered Jul 02 '16 at 14:54

Ramsay Domloge

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3
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1

Good approach, thank you. But it does not show the output as 07. – Emrah Akgül Jul 02 '16 at 15:00
What String input are you trying to decrypt? – Ramsay Domloge Jul 02 '16 at 15:05
All I want to do is xor A2 with A5 and get 07 as string. That's it. – Emrah Akgül Jul 02 '16 at 15:09
1

OK, I've updated the code to work as you describe. Tests out OK – Ramsay Domloge Jul 02 '16 at 15:28

Donald Raab · Answer 2 · 2016-07-03T01:17:47.190

If you are open to using a third party library, the following solution works using Eclipse Collections:

MutableList<String> decryptedPasswordInPairs = Lists.mutable.with("A2")
    .collectWith((oddPair, A5) ->
            CharAdapter.adapt(oddPair)
                .injectIntoWithIndex(
                    ByteLists.mutable.empty(),
                    (bytes, character, index) ->
                        bytes.with((byte) (character ^ A5.charAt(index % A5.length()))))
                .makeString(""), "A5");
decryptedPasswordInPairs.each(System.out::println);

I used the following imports:

import org.eclipse.collections.api.list.MutableList;
import org.eclipse.collections.impl.factory.Lists;
import org.eclipse.collections.impl.factory.primitive.ByteLists;
import org.eclipse.collections.impl.string.immutable.CharAdapter;

By using the primitive ByteList available in Eclipse Collections, I was able to avoid boxing. The method makeString which is available on ByteList allows you to control the separator. Passing in an empty String gets rid of the comma that you get using toString on a List. The class CharAdapter provides a set of char based protocols around String including the method I used here named injectIntoWithIndex.

The code can also be broken into smaller steps to make it slightly less dense and potentially easier to read.

MutableList<String> decryptedPasswordInPairs = Lists.mutable.with("A2")
    .collect(CharAdapter::adapt)
    .collectWith((oddPair, A5) ->
        oddPair.injectIntoWithIndex(
            ByteLists.mutable.empty(),
            (bytes, character, index) ->
                bytes.with((byte) (character ^ A5.charAt(index % A5.length())))), "A5")
    .collectWith(ByteList::makeString, "");
decryptedPasswordInPairs.each(System.out::println);

Note: I am a committer for Eclipse Collections.

score 0 · Answer 3 · edited Jul 02 '16 at 14:43

0

result is displayed as [0,7] instead of 07

This is because you are printing the list as a String.

You could get rid of that doing something like:

for (Byte b : sample) {
    System.out.print(b);
}

Or maybe, but not very elegant:

String result = sample.toString().replace(",", "").replace("[", "").replace("]", "");
decryptedPasswordInPairs.add(result);

edited Jul 02 '16 at 14:43

bcsb1001

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answered Jul 02 '16 at 14:27

ΦXocę 웃 Пepeúpa ツ

47,427
17
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Is there a safe solution for line: String result = sample.toString(); ? I mean is there another toString method for this? – Emrah Akgül Jul 02 '16 at 14:31
1

string is defined in the object class and the list implements it like that... I would use the for loop and some kind of string builder if any is need it :) – ΦXocę 웃 Пepeúpa ツ Jul 02 '16 at 14:32

score -1 · Answer 4 · edited Jun 20 '20 at 09:12

-1

Convert your list into array like below. Thanks @ bcsb1001

T[] toArray(T[] a)

Returns an array containing all of the elements in this list in proper sequence (from first to last element); the runtime type of the returned array is that of the specified array. If the list fits in the specified array, it is returned therein. Otherwise, a new array is allocated with the runtime type of the specified array and the size of this list.

Byte[] byteArray = sample.toArray(new Byte[0])

after that , You can create String from Byte array.

Use this constructor:

String str= new String(byteArray, "UTF-8");

edited Jun 20 '20 at 09:12

Community

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answered Jul 02 '16 at 14:32

Sundararaj Govindasamy

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Is there a way to convert "List sample" to byte Array? – Emrah Akgül Jul 02 '16 at 14:34
2

Sorry that did not work. It says cannot resolve constructor. I guess there is not such constructor. – Emrah Akgül Jul 02 '16 at 14:37
1

Since you've linked the documentation: Show us the constructor, which accepts an `Object[]` ... – Tom Jul 02 '16 at 14:40
2

`byte[] != Object[]`. Also, with the edit, you can't cast an `Object[]` to `Byte`. It should be `Byte[] byteArray = sample.toArray(new Byte[0]);` – bcsb1001 Jul 02 '16 at 14:44
@bcsb1001 one should prefer `sample.toArray(new Byte[sample.size()])` to avoid creating two arrays, instead of one. – Tom Jul 02 '16 at 14:52
1

Still it did not work. I guess there is a difference between Byte [] and byte[]. – Emrah Akgül Jul 02 '16 at 14:52
1

you can refer http://stackoverflow.com/questions/12944377/how-to-convert-byte-to-byte-and-the-other-way-around – Sundararaj Govindasamy Jul 02 '16 at 14:59
This is definitely wrong. It can't work like that because `Byte[] != Byte[]`. The String class has no constructor that accepts `byte[]`. And since autoboxing of arrays is not possible, this answer is wrong. – Easterwood Mar 29 '23 at 08:33

Convert List to string

4 Answers4