Develop Reference

How to replace String with unicode characters

I intend to replace strings with normal strings into split strings, but there is a difference in length between the normal strings which amounts to 62 and the split length turns out to be 117, so when we write the 'a' button it doesn't change to '𝕒' is there another way of writing replace string easier?
public static String doublestruck(String input){
String normal = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
String split = "𝟘𝟙𝟚𝟛𝟜𝟝𝟞𝟟𝟠𝟡𝕒𝕓𝕔𝕕𝕖𝕗𝕘𝕙𝕚𝕛𝕟𝕠𝕡𝕢𝕣𝕤𝕥𝕦𝕧𝕨𝕩𝕪𝕫𝔸𝔹ℂ𝔻𝔼𝔽𝔾ℍ𝕀𝕁𝕂𝕃𝕄ℕ𝕆ℙℚℝ𝕊𝕋𝕌𝕍𝕎𝕏𝕐ℤ";
String output = "";
char letter;
for(int i = 0; i < input.length(); i++){
letter = input.charAt(i);
int a = normal.indexOf(letter);
output += (a != -1) ? split.charAt(a):letter;
}
return new StringBuilder(output).toString();
}

The letters like 𝟜 (U+1D7DC) are not in the Basic Multilingual Pane and thus take up two char values in Java.
Instead of charAt you need to use codePointAt and to find the correct offset you need to use offsetByCodePoint instead of directly using the same index. So split.charAt(a) needs to be replaced by split.codePointAt(spli.offsetByCodePoint(0, a)).

Encode/decode two strings into one string, and back to two

This is an interview question. I was thinking of a solution in java. This questions seems very simple, is there a catch here?
I was thinking of the following solution:
string1 + 1*hash(String1) + string2 + 2*hash(String2).
If I concat strings like this, then I can decode them as well easily into 2 separate strings.
Am I missing something in the question?

Encode:
String encoded = new JsonArray().add(str1).add(str2).toString();
Decode:
JsonArray arr = JsonArray.readFrom(encoded);
String str1 = arr.get(0).asString();
String str2 = arr.get(1).asString();
Here I use minimal-json lib, but it's pretty similar with any other JSON library as well.
Note that it's usually a bad idea to invent new formats of encoding the information into the string as you have plenty of existing ones (xml, json, yaml, etc.) which already solved all the possible issues like symbol escaping and exception handling.

To encode:
String encoded = ""+str1.length()+"/"+str1+str2;
To decode:
String[] temp = encoded.split("/", 2);
int length1 = Integer.parseInt(temp[0]);
String str1 = temp[1].substring(0, length1);
String str2 = temp[1].substring(length1);
Explanation:
The encoded string is in the form "<number>/<str1><str2>". When you call split(regex, limit) the size of the resulting array will be at most limit, considering only the first matches of regex. Thus even if your strings contain the character / you can be sure that the resulting array will be {"<number>", "<str1><str2>"}.
the substring(begin, end) return a string starting at begin inclusive and ending at end exclusive, giving you a resulting substring of end-begin length. Since you are calling it with values(0, str1.length()) what you get is exactly str1. The last call will return a substring from str1.length(), which is also the index of the first character of str2, to the end of the string (which is the end of str2).
Reference: String javadoc page

One way is to use the length of the first string.
// encode
String concat = string1 + string2;
// decode
String str1 = concat.substring( 0, string1.length() );
String str2 = concat.substring( string1.length(), concat.length() );
Another way is to use a delimiter. But the delimiter character should not be included in any of the strings to be joined.
// encode
String concat = "hello" + "`" + "world!";
// decode
String[] decoded = concat.split("`");
String str1 = decoded[0];
String str2 = decoded[1];

How do I recognize a character such as "ç" as a letter?

I have an array of bytes that contains a sentence. I need to convert the lowercase letters on this sentence into uppercase letters. Here is the function that I did:
public void CharUpperBuffAJava(byte[] word) {
for (int i = 0; i < word.length; i++) {
if (!Character.isUpperCase(word[i]) && Character.isLetter(word[i])) {
word[i] -= 32;
}
}
return cchLength;
}
It will work fine with sentences like: "a glass of water". The problem is it must work with all ANSI characters, which includes "ç,á,é,í,ó,ú" and so on. The method Character.isLetter doesn't work with these letters and, therefore, they are not converted into uppercase.
Do you know how can I identify these ANSI characters as a letter in Java?
EDIT
If someone wants to know, I did method again after the answers and now it looks like this:
public static int CharUpperBuffAJava(byte[] lpsz, int cchLength) {
String value;
try {
value = new String(lpsz, 0, cchLength, "Windows-1252");
String upperCase = value.toUpperCase();
byte[] bytes = upperCase.getBytes();
for (int i = 0; i < cchLength; i++) {
lpsz[i] = bytes[i];
}
return cchLength;
} catch (UnsupportedEncodingException e) {
return 0;
}
}

You need to "decode" the byte[] into a character string. There are several APIs to do this, but you must specify the character encoding that is use for the bytes. The overloaded versions that don't use an encoding will give different results on different machines, because they use the platform default.
For example, if you determine that the bytes were encoded with Windows-1252 (sometimes referred to as ANSI).
String s = new String(bytes, "Windows-1252");
String upper = s.toUpperCase();

Convert the byte array into a string, supporting the encoding. Then call toUpperCase(). Then, you can call getBytes() on the string if you need it as a byte array after capitalizing.

Can't you simply use:
String s = new String(bytes, "cp1252");
String upper = s.toUpperCase(someLocale);

Wouldn't changing the character set do the trick before conversion? The internal conversion logic of Java might work fine. Something like http://www.exampledepot.com/egs/java.nio.charset/ConvertChar.html, but use ASCII as the target character set.

I am looking at this table:
http://slayeroffice.com/tools/ascii/
But anything > 227 appears to be a letter, but to make it upper case you would subtract 27 from the ASCII value.

How to Convert ASCII to Hex

How can I convert ASCII values to hexadecimal and binary values (not their string representation in ASCII)? For example, how can I convert the decimal value 26 to 0x1A?
So far, I've tried converting using the following steps (see below for actual code):
Converting value to bytes
Converting each byte to int
Converting each int to hex, via String.toString(intValue, radix)
Note: I did ask a related question about writing hex values to a file.
Clojure code:
(apply str
(for [byte (.getBytes value)]
(.replace (format "%2s" (Integer/toString (.intValue byte) 16)) " " "0")))))
Java code:
Byte[] bytes = "26".getBytes();
for (Byte data : bytes) {
System.out.print(String.format("%2s", Integer.toString(data.intValue(), 16)).replace(" ", "0"));
}
System.out.print("\n");

Hexadecimal, decimal, and binary integers are not different things -- there's only one underlying representation of an integer. The one thing you said you're trying to avoid -- "the ASCII string representation" -- is the only thing that's different. The variable is always the same, it's just how you print it that's different.
Now, it's not 100% clear to me what you're trying to do. But given the above, the path is clear: if you've got a String, convert it to an int by parsing (i.e., using Integer.parseInt()). Then if you want it printed in some format, it's easy to print that int as whatever you want using, for example, printf format specifiers.

If you actually want hexadecimal strings, then (format "%02X" n) is much simpler than the hoops you jump through in your first try. If you don't, then just write the integer values to a file directly without trying to convert them to a string.
Something like (.write out (read-string string-representing-a-number)) should be sufficient.

Here are your three steps rolled up into one line of clojure:
(apply str (map #(format "0x%02X " %) (.getBytes (str 42))))
convert to bytes (.getBytes (str 42))
no actual need for step 2
convert each byte to a string of characters representing it in hex
or you can make it look more like your steps with the "thread last" macro
(->> (str 42) ; start with your value
(.getBytes) ; put it in an array of bytes
(map #(format "0x%02X " %)) ; make hex string representation
(apply str)) ; optionally wrap it up in a string

static String decimalToHex(String decimal, int minLength) {
Long n = Long.parseLong(decimal, 10);
// Long.toHexString formats assuming n is unsigned.
String hex = Long.toHexString(Math.abs(n), 16);
StringBuilder sb = new StringBuilder(minLength);
if (n < 0) { sb.append('-'); }
int padding = minLength - hex.length - sb.length();
while (--padding >= 0) { sb.append('0'); }
return sb.append(hex).toString();
}

//get Unicode for char
char theChar = 'a';
//use this to go from i`enter code here`nt to Unicode or HEX or ASCII
int theValue = 26;
String hex = Integer.toHexString(theValue);
while (hex.length() < 4) {
hex = "0" + hex;
}
String unicode = "\\u" + (hex);
System.out.println(hex);

Get int from String, also containing letters, in Java

How can I get the int value from a string such as 423e - i.e. a string that contains a number but also maybe a letter?
Integer.parseInt() fails since the string must be entirely a number.

Replace all non-digit with blank: the remaining string contains only digits.
Integer.parseInt(s.replaceAll("[\\D]", ""))
This will also remove non-digits inbetween digits, so "x1x1x" becomes 11.
If you need to confirm that the string consists of a sequence of digits (at least one) possibly followed a letter, then use this:
s.matches("[\\d]+[A-Za-z]?")

The NumberFormat class will only parse the string until it reaches a non-parseable character:
((Number)NumberFormat.getInstance().parse("123e")).intValue()
will hence return 123.

Unless you're talking about base 16 numbers (for which there's a method to parse as Hex), you need to explicitly separate out the part that you are interested in, and then convert it. After all, what would be the semantics of something like 23e44e11d in base 10?
Regular expressions could do the trick if you know for sure that you only have one number. Java has a built in regular expression parser.
If, on the other hands, your goal is to concatenate all the digits and dump the alphas, then that is fairly straightforward to do by iterating character by character to build a string with StringBuilder, and then parsing that one.

You can also use Scanner :
Scanner s = new Scanner(MyString);
s.nextInt();

Just go through the string, building up an int as usual, but ignore non-number characters:
int res = 0;
for (int i=0; i < str.length(); i++) {
char c = s.charAt(i);
if (c < '0' || c > '9') continue;
res = res * 10 + (c - '0');
}

Perhaps get the size of the string and loop through each character and call isDigit() on each character. If it is a digit, then add it to a string that only collects the numbers before calling Integer.parseInt().
Something like:
String something = "423e";
int length = something.length();
String result = "";
for (int i = 0; i < length; i++) {
Character character = something.charAt(i);
if (Character.isDigit(character)) {
result += character;
}
}
System.out.println("result is: " + result);