I have hexadecimal String eg. "0x103E" , I want to convert it into integer.
Means String no = "0x103E";
to int hexNo = 0x103E;
I tried Integer.parseInt("0x103E",16); but it gives number format exception.
How do I achieve this ?
You just need to leave out the "0x" part (since it's not actually part of the number).
You also need to make sure that the number you are parsing actually fits into an integer which is 4 bytes long in Java, so it needs to be shorter than 8 digits as a hex number.
No need to remove the "0x" prefix; just use Integer.decode instead of Integer.parseInt:
int x = Integer.decode("0x103E");
System.out.printf("%X%n", x);
Related
I am trying to convert UUID which is coming as string to Big Integer but it's failing every time with Number Format exception as it need String Decimal as parameter. Is there any way we can achieve this.
String x = "6CFAFD0DA976088FE05400144FFB4B37";
I tried with radix also but output is different.
BigInteger big = new BigInteger(x, 0);
System.out.println(big);
Any help is appreciated, TIA.
You are supposed to be using radix 16 as your string has alphanumeric values from 0-9 and A-F, set value 16 in radix as you have hexadecimal string.
String x = "6CFAFD0DA976088FE05400144FFB4B37";
BigInteger big = new BigInteger(x, 16);
System.out.println(big);
OUTPUT
144859830291446118078300087367740640055
You need to set radix value to 16.
For hexadecimal String you need to define the radix value as 16
Well, I have a string in my Java code that needs to be converted into an integer with padding of 10.
Ex. Consider this is the string... Str = "52112"
I need to convert this string into an integer and the result should be like "0000052112". The result should be an integer. Can anyone help me with this, please?
As far as I know you cannot have an integer typed variable with leading zeros. You can pad the number with zeros but then it will become a String.
Take a look at:
http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#leftPad(java.lang.String,%20int)
In order to conform to the signature you have to convert the number to a string first, but that is no great deal.
The leading zeroes have no meaning if the data type you require is an Integer (or any other numeric type). If on the other hand you need a String with leading zeroes, you can use this (works only if required string length is >= number of digits of the number you want to pad) :
String myNumber = Integer.toString(42);
String myNumberWithLeadingZeroes = "0000000000" + myNumber;
// 10 zeroes if you need a string of length 10 in the end
myNumberWithLeadingZeroes = myNumberWithLeadingZeroes.substring(myNumber.length());
I have no problem in writing this: (it also doesn't give any errors)
int hex = 0xFFFFFFFF;
The integer has to have an alpha value also!
But when I try to do this:
Integer hex = Integer.parseInt("0xFFFFFFFF");
// Or I try this:
Integer hex = Integer.parseInt("FFFFFFFF");
I get a java.lang.NumberFormatException: For input string: "0xFFFFFFFF" thrown!
Why is this not working?
I'm guessing that there is some other way to parse hex integers that I am just not realizing, and I really need to be able to parse hex from string for a program I am making.
There is actually a separate function where you can define the radix:
https://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#parseInt(java.lang.String,%20int)
Integer.parseInt(x) simply calls Integer.parseInt(x, 10), so you have to specify a radix 10 number.
In order to parse a hex string, you would simply have to use the following (note that the 0x prefix is not allowed):
Integer.parseInt("FFFFFFF", 16);
I have this for example:
0 10000101 00111100000000000000000
and want to convert this to decimal number.
So far, I already have the code to get the exponent part:
String[]hex={"0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"};
String[]binary={"0000","0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};
String userInput="429E0000";
String result="";
for(int i=0;i<userInput.length();i++)
{
char temp=userInput.charAt(i);
String temp2=""+temp+"";
for(int j=0;j<hex.length;j++)
{
if(temp2.equalsIgnoreCase(hex[j]))
{
result=result+binary[j];
}
}
}
System.out.println(result);
int exponent = Integer.parseInt(result.substring(1,9),2)-127;
System.out.println(exponent);
Is there any in-built command in Java?
Yes, there is a built-in command, intBitsToFloat converts a 32-bit int to a float. You only have to parse your input as an int, the easier way - if your input is in hexadecimal format - would be to directly use base 16 in Integer.parseInt(), then intBitsToFloat converts that bit pattern to a float.
The Integer.ParseInteger(str, radix) will convert a binary digit string to an int ... if you use two as the radix.
However, Daniels answer gives you a better approach that avoids the need for any intermediate string representation of the number.
How can I convert an int number from decimal to binary. For example:
int x=10; // radix 10
How can I make another integer has the binary representation of x, such as:
int y=1010; // radix 2
by using c only?
An integer is always stored in binary format internally -- saying that you want to convert int x = 10 base 10 to int y = 1010 base 2 doesn't make sense. Perhaps you want to convert it to a string representing the binary format of the integer, in which case you can use Integer.toBinaryString.
First thing you should understand is that a value is an abstract notion, that is not bounded to any representation. For example, if you have 20 apples, the number of apples will be the same regardless of the representation. So, dec("10") == bin("1010").
The value of an int reffers to this abstract notion of value, and it does not have any form until you with to print it. This means that the notion of base is important only for conversions from string to int and back.
String s = Integer.toBinaryString(10);
http://download.oracle.com/javase/1.5.0/docs/api/java/lang/Integer.html
Whether it's binary or decimal doesn't really have anything to do with the integer itself. Binary or decimal is a property of a physical representation of the integer, i.e. a String. Thus, the methods you should look at are Integer.toString() and Integer.valueOf() (the versions that take a radix parameter).
BTW, internally, all Java integers are binary, but literals in the source code are decimal (or octal).
Your question is a bit unclear but I'll do my best to try to make sense of it.
How can I make another integer has the binary representation of x such as: int y=1010 radix 2?
From this it looks like you wish to write a binary literal in your source code. Java doesn't support binary integer literals. It only supports decimal, hexadecimal and octal.
You can write your number as a string instead and use Integer.parseInt with the desired radix:
int y = Integer.parseInt("1010", 2);
But you should note that the final result is identical to writing int y = 10;. The integer 10 that was written as a decimal literal in the source code is identical in every way to one which was parsed from the binary string "1010". There is no difference in their internal representation if they are both stored as int.
If you want to convert an existing integer to its binary representation as a string then you can use Integer.toBinaryString as others have already pointed out.
Both integers will have the same interior representation, you can however display as binary via Integer.toBinaryString(i)
Use Integer.toBinaryString()
String y = Integer.toBinaryString(10);
Converting an integer to another base (string representation):
int num = 15;
String fifteen = Integer.toString(num, 2);
// fifteen = "1111"
Converting the string back into an integer
String fifteen = "1111";
int num = Integer.valueOf(fifteen, 2);
// num = 15
This covers the general case for any base. There's no way to explicitly assign an integer as binary (only decimal, octal, and hexadecimal)
int x = 255; // decimal
int y = 0377; // octal (leading zero)
int z = 0xFF; // hex (prepend 0x)