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Converting a hexadecimal string to integer value without pre-made syntax

I'm taking a computer organization class in college. I was tasked with writing a java program that takes a user-inputted string, calls a function that converts said string into a hexadecimal integer, and then outputs the results.
The kicker is that I can't use any existing syntax to do this. for example, Integer.parseInt(__,16) or printf. It all neds to be hardcoded.
Now I'm not asking you to do my homework for me, just wanting to be put in the right direction.
So far, I've made this but can't seem to get the method created right:
import java.util.*;
public class Demo_Class
{
public static void main(String[] args)
{
Scanner AI = new Scanner(System.in);
String str;
System.out.println("Please input a hexadecimal number: ");
str = AI.nextLine();
converter(str);
}
public static int converter(String in)
{
String New = new String();
for(int i = 0; i<= in.length(); i++)
{
New += in.charAt(i);
System.out.println(New + 316);
}
return 0;
}
}

Consider this, lets says you have the hex value 1EC which in hex digits would be 1, E, C. In decimal they would be 1, 14, 12.
so set sum = 0.
sum = sum*16 + 1. sum is now 1
sum = sum*16 + 14 sum is now 30
sum = sum*16 + 12 sum is now 492
So 492 is the answer.
If you have a string of 1EC you need to convert to characters and then convert those characters to the decimal equivalent of hex values.
Try this on paper until you get the feel and then code it. You can check your results using the Integer method you mentioned.

#WJS gave a good hint, I'd just like to add that the charAt() returns the char, which is encoded in ASCII.
As you can see in the ASCII table, the characters A-F have decimal values from 65 to 70, while 0-9 go from 48 to 57 so you'll need to use them to convert the ASCII characters to their intended value.
To do so, you can either get the decimal value of a character by casting to short like short dec = (short)in.charAt(i);, or directly use the characters like char current = in.charAt(i) - 'A'.
With this in mind, all that's left is some calculation, I'll leave that as the homework. :)
Also:
you are looping one character more than needed, change the i <= in.length() to i < in.length(), since it's going from 0
I don't know what that 316 "magic number" is, if it does mean something, declare a variable with a meaningful name, like:
final int MEANINGFUL_NAME = 316;

The readChar() method displays japanese character

I'm trying to write a code that pick-up a word from a file according to an index entered by the user but the problem is that the method readChar() from the RandomAccessFile class is returning japanese characters, I must admit that it's not the first time that I've seen this on my lenovo laptop , sometimes on some installation wizards I can see mixed stuff with normal characters mixed with japanese characters, do you think it comes from the laptop or rather from the code?
This is the code:
package com.project;
import java.io.*;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws IOException {
int N, i=0;
char C;
char[] charArray = new char[100];
String fileLocation = "file.txt";
BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
do {
System.out.println("enter the index of the word");
N = Integer.parseInt(buffer.readLine());
if (N!=0) {
RandomAccessFile word = new RandomAccessFile(new File(fileLocation), "r");
do {
word.seek((2*(N-1))+i);
C = word.readChar();
charArray[i] = C;
i++;
}while(charArray[i-1] != ' ');
System.out.println("the word of index " + N + " is: " );
for (char carTemp : charArray )
System.out.print(carTemp);
System.out.print("\n");
}
}while(N!=0);
buffer.close();
}
}
i get this output :
瑯潕啰灰灥敲牃䍡慳獥攨⠩⤍ഊੴ瑯潌䱯潷睥敲牃䍡慳獥攨⠩⤍ഊ੣捯潭浣捡慴琨⡓却瑲物楮湧朩⤍ഊ੣捨桡慲牁䅴琨⡩楮湴琩⤍ഊੳ獵畢扳獴瑲物楮湧木⠠⁳獴瑡慲牴琠⁩楮湤摥數砬Ⱐ⁥敮湤搠⁩楮湤摥數砩⤍ഊੴ瑲物業洨⠩Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index 100 out of bounds for length 100
at Main.main(Main.java:21)

There are many things wrong, all of which have to do with fundamental misconceptions.
First off: A file on your disk - never mind the File interface in Java, or any other programming language; the file itself - does not and cannot store text. Ever. It stores bytes. That is, raw data, as (on every machine that's been relevant for decades, but historically there have been other ways to do it) quantified in bits, which are organized into groups of 8 that are called bytes.
Text is an abstraction; an interpretation of some particular sequence of byte values. It depends - fundamentally and unavoidably - on an encoding. Because this isn't a blog, I'll spare you the history lesson here, but suffice to say that Java's char type does not simply store a character of text. It stores an unsigned two-byte value, which may represent a character of text. Because there are more characters of text in Unicode than two bytes can represent, sometimes two adjacent chars in an array are required to represent a character of text. (And, of course, there is probably code out there that abuses the char type simply because someone wanted an unsigned equivalent of short. I may even have written some myself. That era is a blur for me.)
Anyway, the point is: using .readChar() is going to read two bytes from your file, and store them into a char within your char[], and the corresponding numeric value is not going to be anything like the one you wanted - unless your file happens to be encoded using the same encoding that Java uses natively, called UTF-16.
You cannot properly read and interpret the file without knowing the file encoding. Full stop. You can at best delude yourself into believing that you can read it. You also cannot have "random access" to a text file - i.e., indexing according to a number of characters of text - unless the encoding in question is constant width. (Otherwise, of course, you can't just calculate the distance-in-bytes into the file where a given character of text is; it depends on how many bytes the previous characters took up, which depends on which characters they are.) Many text encodings are not constant width. One of the most popular, which frankly is the sane default recommendation for most tasks these days, is not. In which case you are simply out of luck for the problem you describe.
At any rate, once you know the encoding of your file, the expected way to retrieve a character of text from a file in Java is to use one of the Reader classes, such as InputStreamReader:
An InputStreamReader is a bridge from byte streams to character streams: It reads bytes and decodes them into characters using a specified charset. The charset that it uses may be specified by name or may be given explicitly, or the platform's default charset may be accepted.
(Here, charset simply means an instance of the class that Java uses to represent text encodings.)
You may be able to fudge your problem description a little bit: seek to a byte offset, and then grab the text characters starting at that offset. However, there is no guarantee that the "text characters starting at that offset" make any sense, or in fact can be decoded at all. If the offset happens to be in the middle of a multi-byte encoding for a character, the remaining part isn't necessarily valid encoded text.

char is 16 bits, i.e. 2 bytes.
seek seeks to a byte boundary.
If the file contains chars then they are at even offsets: 0, 2, 4...
The expression (2*(N-1))+i) is even iff i is even; if odd, you are sure to land in the middle of a char, and thus read garbage.
i starts at zero, but you increment by 1, i.e., half a character.
Your seek argument should probably be (2*(N-1+i)).
Alternative explanation: your file does not contain chars at all; for example, you created an ASCII file in which a character is a single byte.
In that case, the error is attempting to read ASCII (an obsolete character encoding) with a readChar function.
But if the file contains ASCII, the purpose of multiplying by 2 in the seek argument is obscure. It apparently serves no useful purpose.

I changed the encoding of the file to UTF-16 and modified the programe in order to display the right indexes, those that represents the beginning of each word, now it works fine, Thank you guys.
import java.io.*;
public class Main {
public static void main(String[] args) throws IOException {
int N, i=0, j=0, k=0;
char C;
char[] charArray = new char[100];
String fileLocation = "file.txt";
BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
DataInputStream in = new DataInputStream(new FileInputStream(fileLocation));
boolean EOF=false;
do {
try {
j++;
C = in.readChar();
if((C==' ')||(C=='\n')){
System.out.print(j+1+"\t");
}
}catch (IOException e){
EOF=true;
}
}while (EOF!=true);
System.out.println("\n");
do {
System.out.println("enter the index of the word");
N = Integer.parseInt(buffer.readLine());
if (N!=0) {
RandomAccessFile word = new RandomAccessFile(new File(fileLocation), "r");
do {
word.seek((2*(N-1+i)));
C = word.readChar();
charArray[i] = C;
i++;
}while(charArray[i-1] != ' ' && charArray[i-1] != '\n');
System.out.print("the word of index " + N + " is: " );
for (char carTemp : charArray )
System.out.print(carTemp);
System.out.print("\n");
i=0;
charArray = new char[100];
}
}while(N!=0);
buffer.close();
}
}

Multiplying ascii code and getting string back

I am trying to convert a string to ascii code and then multiplying that concatenation of the ascii codes by a number.
For example
String message = "Hello";
String result = "";
ArrayList arrayList = new ArrayList();
int temp;
for(int i = 0; i < message.length(); i++){
temp = (int) message.charAt(i);
result = result + String.value(temp).toString();
arrayList.add(String.valueOf(temp).toString());
}
I have tried two different ways, but there is always a catch with each one.
If I just concatenate all the ascii codes together into a string and I get 72101108108111 as my new string, the problem now is how can I get the original string back from this? This is because it is not obvious where each one character code starts and ends and the next one begins.
Another way I tried doing this was to use an array. I would receive |72|101|108|108|111| in an array. Obviously the codes are split here, but if I wanted to multiply this whole array (all the numbers as one number) by a number and then how would I get the array back together?
These are two different ways I have thought to solve this, but I have no idea how to get the string back out of these if I multiply the ascii by a number.

You don't need to modify the original string nor the ascii code string. Just have them both there, then whenever you need to get the numerical value of the string, just use X.valueOf(...)** method. Example,
final String message = "Hello";
final String result = "72101108108111";
long value = Long.valueOf(result);
If you do not want to store the two strings, then you should go with the array method. To get a numerical value, you simply concatenate all the strings in the array into one and use the X.valueOf(..) method.
And to get back the original string, use Integer.valueOf(...) on each string in the array, then cast each one to char.
System.out.println((char)Integer.valueOf("111").intValue());
** Note by X.valueOf(..), X doesn't have to be Long or Integer as I have shown. As you mentioned the value can get really large so BigInteger should be prefered above others

Exception while Parsing Hexadecimal to Short in Java

In file data.hex, the data is given in hexadecimal form where the first digit of hexadecimal number is always less than 8. Eg.
01FC 04BF 04C0 04C1 04C2 24C3 04C4 34C5 ...
To parse this file and store the values in shrt[] array i have written this code
void read_hex_short(String filename, short[] shrt, int x, int y) throws Exception
{
String str;
Scanner s=new Scanner(new BufferedReader(new FileReader(filename)));
for(int i=0;i<height*width;i++)
{
str= s.next(); // i have tried str="0x"+s.next() but it didn't work
image[i]=(short)Integer.parseInt(str);
}
s.close();
}
But i am getting NumberFormatException which arises while passing the first string i.e. 01FC only.
How can i parse these hexadecimal values and store them in the shrt[] array?

You should be using Integer.parseInt(str, 16) to tell it to use hex.
You should also be aware that any values greater than 0x7FFF will end up being negative in your array: Java doesn't have any unsigned numeric types (unless you count char).

You can use Short.parseShort(str, 16) to parse HEX. This will avoid the need to cast it to short from an int.

Give parseInt method a radix parameter. Like this:
Integer.parseInt(str, 16)

java convert to int

I am trying to convert to int like this, but I am getting an exception.
String strHexNumber = "0x1";
int decimalNumber = Integer.parseInt(strHexNumber, 16);
Exception in thread "main" java.lang.NumberFormatException: For input string: "0x1"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:458)
It would be a great help if someone can fix it.
Thanks.

That's because the 0x prefix is not allowed. It's only a Java language thing.
String strHexNumber = "F777";
int decimalNumber = Integer.parseInt(strHexNumber, 16);
System.out.println(decimalNumber);
If you want to parse strings with leading 0x then use the .decode methods available on Integer, Long etc.
int value = Integer.decode("0x12AF");
System.out.println(value);

Sure - you need to get rid of the "0x" part if you want to use parseInt:
int parsed = Integer.parseInt("100", 16);
System.out.println(parsed); // 256
If you know your value will start with "0x" you can just use:
String prefixStripped = hexNumber.substring(2);
Otherwise, just test for it:
number = number.startsWith("0x") ? number.substring(2) : number;
Note that you should think about how negative numbers will be represented too.
EDIT: Adam's solution using decode will certainly work, but if you already know the radix then IMO it's clearer to state it explicitly than to have it inferred - particularly if it would surprise people for "035" to be treated as octal, for example. Each method is appropriate at different times, of course, so it's worth knowing about both. Pick whichever one handles your particular situation most cleanly and clearly.

Integer.parseInt can only parse strings that are formatted to look just like an int. So you can parse "0" or "12343" or "-56" but not "0x1".

You need to strip off the 0x from the front of the string before you ask the Integer class to parse it. The parseInt method expects the string passed in to be only numbers/letters of the specified radix.

try using this code here:-
import java.io.*;
import java.lang.*;
public class HexaToInteger{
public static void main(String[] args) throws IOException{
BufferedReader read =
new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the hexadecimal value:!");
String s = read.readLine();
int i = Integer.valueOf(s, 16).intValue();
System.out.println("Integer:=" + i);
}
}

Yeah, Integer is still expecting some kind of String of numbers. that x is really going to mess things up.

Depending on the size of the hex, you may need to use a BigInteger (you can probably skip the "L" check and trim in yours ;-) ):
// Convert HEX to decimal
if (category.startsWith("0X") && category.endsWith("L")) {
category = new BigInteger(category.substring(2, category.length() - 1), 16).toString();
} else if (category.startsWith("0X")) {
category = new BigInteger(category.substring(2, category.length()), 16).toString();
}