Java Hexadecimal to Decimal conversion: Custom Logic - java
I am trying to figure out how to convert hex into a string and integer so I can manipulate an RGB light on my arduino micro-controller through it's serialport. I found a good example on the java website, but I'm having a difficult time understanding some of the methods and I am getting hung up. I could easily just copy-paste this code and have it work but I want to fully understand it. I will add comments to my understandings and hopefully someone can provide some feedback.
public class HexToDecimalExample3{
public static int getDecimal(String hex){ //this is the function which we will call later and they are declaring string hex here. Can we declare string hex inside the scope..?
String digits = "0123456789ABCDEF"; //declaring string "digits" with all possible inputs in linear order for later indexing
hex = hex.toUpperCase(); //converting string to uppercase, just "in case"
int val = 0; //declaring int val. I don't get this part.
for (int i = 0; i < hex.length(); i++) //hex.length is how long the string is I think, so we don't finish the loop until all letters in string is done. pls validate this
{
char c = hex.charAt(i); //char is completely new to me. Are we taking the characters from the string 'hex' and making an indexed array of a sort? It seems similar to indexOf but non-linear? help me understand this..
int d = digits.indexOf(c); //indexing linearly where 0=1 and A=11 and storing to an integer variable
val = 16*val + d; //How do we multiply 16(bits) by val=0 to get a converted value? I do not get this..
}
return val;
}
public static void main(String args[]){
System.out.println("Decimal of a is: "+getDecimal("a")); //printing the conversions out.
System.out.println("Decimal of f is: "+getDecimal("f"));
System.out.println("Decimal of 121 is: "+getDecimal("121"));
}}
To summerize the comments, it's primarily the char c = hex.charAt(i); AND the val = 16*val + d; parts I don't understand.
Ok, let's go line for line
public static int getDecimal(String hex)
hex is the parameter, it needs to be declared there, so you can pass a String when you call the function.
String digits = "0123456789ABCDEF";
Yes, this declares a string with all characters which can occur in a hexadecimal number.
hex = hex.toUpperCase();
It converts the letters in the hex-String to upper case, so that it is consistent, i.e. you always have F and never f, no matter which is being input.
int val = 0;
This is the variable where the corresponding decimal value will later be in. We will do our calculations with this variable.
for (int i = 0; i < hex.length(); i++)
hex.length() is the number of characters in the hex-String provided. We execute the code inside this for loop once per character.
char c = hex.charAt(i);
Yes, char represents a single character. We retrieve the character from the hex-String at index i, so in the first iteration it is the first character, in the second iteration the second character and so on.
int d = digits.indexOf(c);
We look which index the character has in the digit-String. In that way we determine the decimal representation of this specific digit. Like 0-9 stay 0-9 and F becomes a 15.
val = 16*val + d;
Let's think about what we have to do. We have the decimal value of the digit. But in hexadecimal we have this digit at a specific position with which it gets multiplied. Like the '1' in '100' is actually not a 1, but 100 * 1 because it is at this position.
10 in hexadecimal is 16 in decimal, because we have 1 * 16. Now the approach here is a little bit complicated. val is not uninitialized. val is 0 at the beginning and then contains the cumulated values from the previous iterations. Since the first character in the String is the highest position we don't know directly with what we have to multiply, because we don't know how many digits the number has (actually we do, but this approach doesn't use this). So we just add the digit value to it. In the consecutive iterations it will get multiplied by 16 to scale it up to the corresponding digit base value. Let me show you an example:
Take 25F as hex number. Now the first iteration takes the 2 and converts it to a 2 and adds it to val. The 16 * val resolves to 0 so is not effective in the first time.
The next iteration multiplies the 2 with 16 and takes the 5 (converted to 5) and adds it to val. So now we have (I split it mathematically so you understand it):
2 * 16 + 5
Next we get the F which is decimal 15. We multiply val by 16 and add the 15.
We get 2 * 256 + 5 * 16 + 16 (* 1), which is actually how you calculate the decimal value of this hex value mathematically.
Another possibility to compute val is:
val += Math.pow(16, hex.length() - i - 1) * d;
Related
Getting letter from integer index
I wish to have a java method which gives me, index given, a corresponding letter set excel like, so:
258 => IZ (last index)
30 => AD
120 => DR
56 => BD
First method gives correct output, but it's very dumb and I don't like that.
I tried to build a second method that involves a bit of thinking.
I already saw methods using String Builder or something else like this one
but I tried to build a method myself aka betterGetColumnName.
better 258 => IHGFEDCBAX (not ok)
better 30 => AD (OK, 2nd alphabet round it's ok)
better 120 => DCBAP (not ok)
better 56 => BAD (seems like 3rd alphabet round breaks my logic)
public String getColumnName(int index){
String[] letters = {
"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R",
"S","T","U","V","W","X","Y","Z","AA","AB","AC","AD","AE","AF","AG","AH",
"AI","AJ","AK","AL","AM","AN","AO","AP","AQ","AR","AS","AT","AU","AV",
"AW","AX","AY","AZ","BA","BB","BC","BD","BE","BF","BG","BH","BI","BJ",
"BK","BL","BM","BN","BO","BP","BQ","BR","BS","BT","BU","BV","BW","BX",
"BY","BZ","CA","CB","CC","CD","CE","CG","CH","CI","CJ","CK","CL","CM",
"CN","CO","CP","CQ","CR","CS","CT","CU","CV","CW","CX","CY","CZ","DA",
"DB","DC","DD","DF","DG","DH","DI","DJ","DK","DL","DM","DN","DO","DP",
"DQ","DR","DS","DT","DU","DV","DW","DX","DY","DZ","EA","EB","EC","ED",
"EE","EF","EG","EH","EI","EJ","EK","EL","EM","EN","EO","EP","EQ","ER",
"ES","ET","EU","EV","EW","EX","EY","EZ","FA","FB","FC","FD","FE","FF",
"FG","FH","FI","FJ","FK","FL","FM","FN","FO","FP","FQ","FR","FS","FT",
"FU","FV","FW","FX","FY","FZ","GA","GB","GC","GD","GE","GF","GG","GH",
"GI","GJ","GK","GL","GM","GN","GO","GP","GQ","GR","GS","GT","GU","GV",
"GW","GX","GY","GZ","HA","HB","HC","HD","HE","HF","HG","HH","HI","HJ",
"HK","HL","HM","HN","HO","HP","HQ","HR","HS","HT","HU","HV","HW","HX",
"HY","HZ","IA","IB","IC","ID","IE","IF","IG","IH","II","IJ","IK","IL",
"IM","IN","IO","IP","IQ","IR","IS","IT","IU","IV","IW","IX","IY","IZ"
};
if (index<=letters.length){
return letters[index-1];
}else{
return null;
}
}
I think I should save how many times I made a full alphabet round, I wouldn't use StringBuilder or else, just char, String and integers because at school we can't upgrade java version (1.5.x) also I think it might be useful for me to understand why is my logic so wrong.
public String betterGetColumnName(int index){
int res=0;
String s = "";
char h='0';
while(index>26){
res=index/26;
h=(char)(res+64);
s+=h;
index -=26;
}
h=(char)(index+64);
s+=h;
return s;
}
You are definitely on the right track, though your logic is a bit off. What you are effectively trying to do is to convert a base 10 integer into a base 26 character. But instead of digits, the converted "number" actually consists of the 26 letters of the alphabet.
The algorithm you want here is to determine each letter of the output by taking the remainder of the input number divided by 26. Then, divide the input by 26 and again inspect the "tens" position to see what letter it is. In the code snippet below, I assume that 1 corresponds to A, 26 corresponds to Z, and 27 to AA. You may shift the indices however you feel is best.
int input = 53;
String output = "";
while (input > 0) {
int num = (input - 1) % 26;
char letter = (char)(num+65);
output = letter + output;
input = (input-1) / 26;
}
System.out.println(output);
BA
Demo
Note: A helpful edit was suggested which uses StringBuilder instead of String to do the concatenations. While this might be more optimal than the above code, it might make it harder to see the algorithm.
Need some explanation about how the hex to decimal algorithm works
The algorithm below converts hex to decimal, but I've confused, how this solution works?
public static int hex2decimal(String s) {
String digits = "0123456789ABCDEF";
s = s.toUpperCase();
int val = 0;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
int d = digits.indexOf(c);
val = 16*val + d;
}
return val;
}
source
I had known only one approach to do that before I found out this one.
I mean, everyone knows:
X*16^Y where X is the number you want to convert and Y is the position for the number (begging from the end to the start).
So, if you want to convert DA145 to decimal would be...
*(5 * 16^0) + (4 * 16^1) + (1 * 16^2) + (10 * 16^3) + (13 * 16^4)*
This algorithm uses the fact that we can calculate 16^Y by repeatedly multiply together 16's and also that we can factor out common multiplications by 16. From your example you would instead end at: 13*16 + 10)*16 + 1)*16 + 4)*16 + 5 I've omitted the leading parentheses, as you see it happens to be that 13 is in effect multiplied with 16 four times.
The algorithm does pretty much the same as you would do it. It takes a String and compares every single char with the given value. This value is evaluated by digits, at which position it stands (e.g. A is in the 10th position, since we start counting with 0). This allows to easily change it e.g. to a 17 digit system instead of hexadezimal. Edit: About the powers of 16, look at #skykings answer.
Need help: Program to convert binary to decimals (Java)
I would like some help with my code. I'm doing a method to convert binary numbers to decimals. This is my code:
public double decimal(double number){
String num=(number+"");
char charac;
double n;
double cont=0;
double exp;
double res=0;
for(int i=num.length()-1;i>=0;i--){
charac=num.charAt(i);//obtains digits 1 by 1 and converts to char
n=(Character)charac;//converts the digit into a number (double)
exp=Math.pow(2, cont);//makes the exponential function to multiply to n
n=n*exp;//multiplies exponential with the digit
res+=n;//adds digit to accumulator
cont++;
}
return res;
}
the problem i'm having is that the numbers get all messed up for whatever reason, like n being assigned 48 in the first loop of the for cycle.
I tried using n as an Int instead and it seemed to be working well but at the second loop of the for cycle it was assigned -2 somehow and that ruined the addition.
Change n=(Character)charac; to use Character.digit(char, int) otherwise you get the ascii value (not the digit value) n=Character.digit(charac, 10);
By assigning char to double, it is infact passing on the ASCII value of that digit e.g. 48 is returned for char 0.
I think this answer from How to convert binary string value to decimal will help you: String c = "110010"; // as binary int decimalValue = Integer.parseInt(c, 2); System.out.println(decimalValue); result: 50
10bit binary number from a string to a byte
Currently I have a 8 bit binary string and I want to shift it left and get a 10bit binary number from it.
(ie 0111001010)
String[] channels = datArray.get(n).split(" ");
byte[] RI = channels[m+1].getBytes();
String s = (Integer.toBinaryString(Integer.parseInt(channels[m+1])));
Example Values:
RI is equal to: [B#4223b758
S is equal to: 1100101
Any help is much appreciated.
Wouldn't this work for you:
String input = "1100101";
int value = Integer.parseInt(input, 2) << 1;
System.out.println(Integer.toBinaryString(value));
Returns:
11001010
Parsed binary string and shifted left (one digit).
The two things that it looks like you are missing from your approach are the ability to specify a radix when parsing a String representing a binary numeral, and the left shift operator.
If you wanted the leading zeros I was surprised to see that there is no built in way to accomplish this, and the current wisdom is this is the optimal way (taken from this discussion)
System.out.println(String.format("%10s", Integer.toBinaryString(value)).replace(' ', '0'));
Which for the example value given would return this:
0011001010
You can use following:
BigInteger i = new BigInteger("00000011" + "00", 2);
int x = i.intValue();
Where "00000011" is your string representation of 8 bit number. The "00" simulates the left shifting..
Can I multiply charAt in Java?
When I try to multiply charAt I received "big" number: String s = "25999993654"; System.out.println(s.charAt(0)+s.charAt(1)); Result : 103 But when I want to receive only one number it's OK . On the JAVA documentation: the character at the specified index of this string. The first character is at index 0. So I need explanation or solution (I think that I should convert string to int , but it seems to me that is unnesessary work)
char is an integral type. The value of s.charAt(0) in your example is the char version of the number 50 (the character code for '2'). s.charAt(1) is (char)53. When you use + on them, they're converted to ints, and you end up with 103 (not 100). If you're trying to use the numbers 2 and 5, yes, you'll have to parse them. Or if you know they're standard ASCII-style digits (character codes 48 through 57, inclusive), you can just subtract 48 from them (as 48 is the character code for '0'). Or better yet, as Peter Lawrey points out elsewhere, use Character.getNumericValue, which handles a broader range of characters.
Yes - you should parse extracted digit or use ASCII chart feature and substract 48:
public final class Test {
public static void main(String[] a) {
String s = "25999993654";
System.out.println(intAt(s, 0) + intAt(s, 1));
}
public static int intAt(String s, int index) {
return Integer.parseInt(""+s.charAt(index));
//or
//return (int) s.charAt(index) - 48;
}
}