I'm learning JAVA and recently I had the same problem with a few training tasks.
I have a some numbers and some of them are starting with 0. I found out that these numbers are octal which means it won't be the number I wanted or it gives me an error (because of the "8" or the "9" because they are not octal digits) after I read it as an int or long...
Until now I only had to work with two digit numbers like 14 or 05.
I treated them as Strings and converted them into numbers with a function that checks all of the String numbers and convert them to numbers like this
String numStr = "02";
if(numStr.startsWith("0")) {
int num = getNumericValue(numStr.charAt(1));
} else {
int num = Integer.parseInt(numStr);
}
Now I have an unkown lot of number with an unknown number of digits (so maybe more than 2). I know that if I want I can use a loop and .substring(), but there must be an easier way.
Is there any way to simply ignore the zeros somehow?
Edit:
Until now I always edited the numbers I had to work with to be Strings because I couldn't find an easier way to solve the problem. When I had 0010 27 09 I had to declare it like:
String[] numbers = {"0010", "27", "09"};
Because if I declare it like this:
int[] numbers = {0010, 27, 09};
numbers[0] will be 8 instead of 10 and numbers[2] will give me an error
Actually I don't want to work with Strings. What I actually want is to read numbers starting with zero as numbers (eg.: int or long) but I want them to be decimal. The problem is that I have a lot of number from a source. I copied them into the code and edited it to be a declaration of an array. But I don't want to edit them to be Strings just to delete the zeros and make them numbers again.
I'm not quite sure what you want to achieve. Do you want to be able to read an Integer, given as String in a 8-based format (Case 1)? Or do you want to read such a String and interpret it as 10-based though it is 8-based (Case 2)?
Or do you simply want to know how to create such an Integer without manually converting it (Case 3)?
Case 1:
String input = "0235";
// Cut the indicator 0
input = input.substring(1);
// Interpret the string as 8-based integer.
Integer number = Integer.parseInt(input, 8);
Case 2:
String input = "0235";
// Cut the indicator 0
input = input.substring(1);
// Interpret the string as 10-based integer (default).
Integer number = Integer.parseInt(input);
Case 3:
// Java interprets this as octal number
int octal = 0235;
// Java interprets this as hexadecimal number
int hexa = 0x235
// Java interprets this as decimal number
int decimal = 235
You can expand Case 1 to a intelligent method by reacting to the indicator:
public Integer convert(final String input) {
String hexaIndicator = input.substring(0, 2);
if (hexaIndicator.equals("0x")) {
return Integer.parseInt(input.substring(2), 16);
} else {
String octaIndicator = input.substring(0, 1);
if (octaIndicator.equals("0")) {
return Integer.parseInt(input.substring(1), 8);
} else {
return Integer.parseInt(input);
}
}
}
Related
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;
OK so I am writing a program that compares the number entered by a user to a computer generated number. Each number is 3 digits. What I'd like to know how to do is compare these integers. So for example if the last two digits of the three digit numbers are the same it will run a block of code. I'm not completely sure how to go about this. Any youtube videos or any links would be much appreciated. I would rather understand than get an answer to my code.
The fastet way would be: Convert the Integer to a String and then split the string at each char. Then you have an Char Array with each digit and you can compare them like whatever you want.
int number = 324;
String number_string = String.valueOf(number); //converts integer to string
char[] digits = number_string.toCharArray(); //converts string to char array
if(digits[2] == 4){ //checks if 3rd digit of the char array is 4
//do something
}
You can get last two digit by modulo(%) operator.when you modulo three digit number by 100 it return last two digit and compare these digit in if condition like,
int num=154; //number of three digit
int rem=num%100;//modulo by 100 return remainder
if(rem==54){ //checking condition
//code
}
else{
//code
}
I suppose that the number entered is a String so first you want to convert it to an integer.
String value = .... <whatever read from the user>
int number = Integer.parseInt(value)
Then, if you like to get information on some of the digits, use modulus, ie:
number%10
will give you the most right integer (ie 127%10 is 7)
number%100
will give you 2 digits (ie 626%100 is 26)
I want to display the length of a 4-digit number displayed by the user, the problem that I'm running into is that whenever I read the length of a number that is 4-digits long but has trailing zeros the length comes to the number of digits minus the zeros.
Here is what I tried:
//length of values from number input
int number = 0123;
int length = (int)Math.log10(number) + 1;
This returns to length of 3
The other thing I tried was:
int number = 0123;
int length = String.valueOf(number).length();
This also returned a length of 3.
Are there any alternatives to how I can obtain this length?
Because int number = 0123 is the equivalent of int number = 83 as 0123 is an octal constant. Thanks to #DavidConrad and #DrewKennedy for the octal precision.
Instead declare it as a String if you want to keep the leading 0
String number = "0123";
int length = number.length();
And then when you need the number, simply do Integer.parseInt(number)
Why is the syntax of octal notation in java 0xx ?
Java syntax was designed to be close to that of C, see eg page 20 at
How The JVM Spec Came To Be keynote from the JVM Languages Summit 2008
by James Gosling (20_Gosling_keynote.pdf):
In turn, this is the way how octal constants are defined in C language:
If an integer constant begins with 0x or 0X, it is hexadecimal. If it
begins with the digit 0, it is octal. Otherwise, it is assumed to be
decimal...
Note that this part is a C&P of #gnat answer on programmers.stackexchange.
https://softwareengineering.stackexchange.com/questions/221797/reasoning-behind-the-syntax-of-octal-notation-in-java
Use a string instead:
String number = "0123";
int length = number.length(); // equals 4
This doesn't work with an int, as the internal representation of 0123 is identical to 123. The program doesn't remember how the value was written, only the actual value.
What you can do is declare a string:
String number = "0123";
int numberLengthWithLeadingZeroes = number.length(); // == 4
int numberValue = Integer.parseInt(number); // == 123
If you really want to include leading 0's you could always store it in an array of of characters
example:
char[] abc = String.valueOf(number).toCharArray();
Then obviously you can figure out the length of the array.
As several people have pointed out already though, integers don't have leading 0's.
String abc=String.format("%04d", yournumber);
for zero-padding with length=4.
Refer to this link .
http://download.oracle.com/javase/7/docs/api/java/util/Formatter.html
this might help u ..
Integers in Java don't have leading zeroes. If you assign the value 0123 to your int or Integer variable, it will be interpreted as an octal constant rather than a decimal one, which can lead to subtle bugs.
Instead, if you want to keep leading zeroes, use a String, e.g.
String number = "0123";
This way you can also measure the length:
String number = "0123";
System.out.println(number.length());
I have a program that I am working on which would effectively convert binary, decimal, or hex numbers to other formats (don't ask why I'm doing this, I honestly don't know). So far I only have a binary - decimal conversion going, and it works fine however whenever the binary number entered is 8 digits or more it crashes.
As far as I can tell when I input the number 10011001 it gets translated to scientific notation and becomes 1.0011001E7 which wouldn't really be a problem, except that the way I am converting the numbers involves creating a string with the same value as the number and breaking it into individual characters. Unfortunately, this means I have a string valued "1.0011001E7" instead of "10011001", so when I cut up the characters I hit the "." and the program doesn't know what to do when I try to make calculations with that. So basically my question comes down to this, how do I force it to use the not-scientific notation version for these calculations?
Thanks for all your help, and here is the code if it helps at all:
//This Splits A Single String Of Digits Into An Array Of Individual Digits
public float[] splitDigits(float fltInput){
//This Declares The Variables
String strInput = "" + fltInput;
float[] digit = new float[strInput.length() - 2];
int m = 0;
//This Declares The Array To Hold The Answer
for (m = 0; m < (strInput.length() - 2); m++){
digit[m] = Float.parseFloat(strInput.substring(m, m + 1)); //Breaks here
}
//This Returns The Answer
return digit;
}
Just use BigDecimal
BigDecimal num = new BigDecimal(fltInput);
String numWithNoExponents = num.toPlainString();
Note here the fltInput will be automatically converted to a double.
So I have a set of base digits like "BCDFGHJKLMNPQRSTVWXZ34679"
how do I convert a value say "D6CN96W6WT" to binary string in Java?
This should work (assuming 0,1 for you binary digits):
// your arbitrary digits
private static final String DIGITS = "BCDFGHJKLMNPQRSTVWXZ34679";
public String base25ToBinary(String base25Number) {
long value = 0;
char[] base25Digits = base25Number.toCharArray();
for (char digit : base25Digits) {
value = value * 25 + DIGITS.indexOf(digit);
}
return Long.toString(value, 2);
}
Off the top of my head, for base-25 strings.
Integer.toString(Integer.valueof(base25str, 25), 2)
Its a little unclear from your question whether you're talking about actual 0-9-Z bases, or a number encoding with an arbitrary list of symbols. I'm assuming the first, if its the later then you're out of luck on built-ins.