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Java scanner nextInt(16) does not accept negative hex values

I need to read in a 32-bit number in hex format. When I enter a negative value, I get an input mismatch exception. Everything works as long as the number is positive (00000000 ~ 7FFFFFFF), but anything negative (80000000 ~ FFFFFFFF) fails.
System.out.println("Enter first number in hexadecimal format: ");
Scanner readX = new Scanner(System.in);
int a = readX.nextInt(16);
I have tried various formats (FFFFFFFF, 0xFFFFFFFF, -FFFFFFFF, -7FFFFFFFF, ~FFFFFFFF) with the same results.
Any ideas? I feel like I must be missing something obvious but I'm completely stumped!

It will fail for the same reason 2147483648 (one more than Integer.MAX_VALUE) will fail for Integer.parseInt: The value is too large. It will not interpret ffffffff or 80000000 as a negative number, but as a large positive number. Those numbers are simply too large to be interpreted as an int.
Scanner.nextInt(int radix) matches a regular expression to see if it could be an int, then it passes it to Integer.parseInt for parsing:
If the next token matches the Integer regular expression defined above then the token is converted into an int value as if by removing all locale specific prefixes, group separators, and locale specific suffixes, then mapping non-ASCII digits into ASCII digits via Character.digit, prepending a negative sign (-) if the locale specific negative prefixes and suffixes were present, and passing the resulting string to Integer.parseInt with the specified radix.
Integer.parseInt will throw a NumberFormatException if it can't be represented as an int:
An exception of type NumberFormatException is thrown if any of the following situations occurs:
The first argument is null or is a string of length zero.
The radix is either smaller than Character.MIN_RADIX or larger than Character.MAX_RADIX.
Any character of the string is not a digit of the specified radix, except that the first character may be a minus sign '-' ('\u002D') or plus sign '+' ('\u002B') provided that the string is longer than length 1.
The value represented by the string is not a value of type int.
You must specify the number to be negative. Try -1 or -80000000 or anything in between; they will work.

System.out.println("Enter first number in hexadecimal format: ");
Scanner readX = new Scanner(System.in);
String a = readX.nextLine().trim();
int result = Integer.parseInt(a, 16);
This will be work for youThen you get value as scientific notation if it's not fit to int.

CharSequence to Integer with multiple +ve and -ve signss

i have learned that to convert charsequence to integer we can use this statement
String cs="123";
int number = Integer.parseInt(cs.toString());
what if
cs = "++-+--25";
will this statement still run and give answer -25 according to string given??

You are end up with a NumberFormatException since ++-+--25 is not a valid integer.
See the docs of parseInt()
Parses the string argument as a signed decimal integer. The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign '-' ('\u002D') to indicate a negative value or an ASCII plus sign '+' ('\u002B') to indicate a positive value. The resulting integer value is returned, exactly as if the argument and the radix 10 were given as arguments to the parseInt(java.lang.String, int) method.
So you are allowed to do
CharSequence cs = "-25"; //gives you -25
and
CharSequence cs = "+25"; //gives you 25
Otherwise ,take necessary steps to face the Exception :)
So know the char Sequence is a valid string just write a simple method to return true or false and then proceed further
public static boolean {
try {
Integer.parseInt(s);
} catch(NumberFormatException e) {
return false; // no boss you entered a wrong format
}
return true; //valid integer
}
Then your code looks like
if(isInteger(cs.toString())){
int number = Integer.parseInt(cs.toString());
// proceed remaining
}else{
// No, Operation cannot be completed.Give proper input.
}

Answer to your question is code will run and throw Exception as "++-+--25" is not a valid int,
java.lang.NumberFormatException: For input string: "++-+--25"

You will get
java.lang.NumberFormatException: For input string: "++-+--25"
Tested example :
CharSequence cs = "++-+--25";
System.out.println("" + Integer.parseInt(cs.toString()));

java.lang.NumberFormatException: For input string: "10.0"

This code must validate input data from the findActions() method:
try {
System.out.println(findActions(lookingArea.substring(0, right)));// always printing valid number string
Integer.parseInt(findActions(lookingArea.substring(0, right)));// checking for number format
}
catch(NumberFormatException exc) {
System.out.println(exc);
}
But I always have java.lang.NumberFormatException: For input string: "*number*"
that is so strange, because checking with System.out.println(findActions(lookingArea.substring(0, right)));,
I get *number* like 10.0

Integer.parseInt doesn't expect the . character. If you're sure it can be converted to an int, then do one of the following:
Eliminate the ".0" off the end of the string before parsing it, or
Call Double.parseDouble, and cast the result to int.
Quoting the linked Javadocs above:
The characters in the string must all be decimal digits, except that
the first character may be an ASCII minus sign '-' ('\u002D') to
indicate a negative value or an ASCII plus sign '+' ('\u002B') to
indicate a positive value.

10.0 is not an integer number. Instead, you can use:
int num = (int) Double.parseDouble(...);

One of the reason could be that your string is too long to convert into Integer type, So you can declare it as Long or Double based on the provided input.
Long l = Long.parseLong(str);

NumberFormat Exception to parse String to Byte?

I am trying to parse the following String to Byte.But it gives me NumberFormat Exception.Can some body tell me what is the solution for this?
Byte.parseByte("11111111111111111111111110000001", 2);

Byte.parseByte() handles binary string as sign-magnitude not as a 2's complement, so the longest length you can have for a byte is 7 bits with a sign.
In other words, to represent -127, you should use:
Byte.parseByte("-111111", 2);
The following throws NumberFormatException:
Byte.parseByte("10000000", 2);
However, the binary literal of -127 is:
byte b = (byte) 0b10000000;
The same behavior is applied to the other parseXXX() methods.

Out of range of byte ie -128 to 127. From parseByte(String s,int radix) javadoc:
public static byte parseByte(String s, int radix)throws NumberFormatException
Parses the string argument as a signed byte in the radix specified by
the second argument. The characters in the string must all be digits,
of the specified radix (as determined by whether Character.digit(char,
int) returns a nonnegative value) except that the first character may
be an ASCII minus sign '-' ('\u002D') to indicate a negative value.
The resulting byte value is returned. An exception of type
NumberFormatException is thrown if any of the following situations
occurs:
The first argument is null or is a string of length zero.
The radix is either smaller than Character.MIN_RADIX or larger than
Character.MAX_RADIX.
Any character of the string is not a digit of the specified radix,
except that the first character may be a minus sign '-' ('\u002D')
provided that the string is longer than length 1.
The value represented by the string is not a value of type byte.
Returns: the byte value represented by the string argument in the
specified radix Throws: NumberFormatException - If the string does not
contain a parsable byte.

from javadocs
An exception of type NumberFormatException is thrown if any of the
following situations occurs:
The first argument is null or is a string of length zero.
The radix is either smaller than Character.MIN_RADIX or larger than Character.MAX_RADIX.
Any character of the string is not a digit of the specified radix, except that the first character may be a minus sign '-' ('\u002D')
provided that the string is longer than length 1.
The value represented by the string is not a value of type byte.
Your value is the second case that is out of range -128 to 127

Value is too large to be parsed in byte
Try this:
new BigInteger("011111111111111111111111110000001", 2).longValue();

String binary to Hex Java

i have code that looks like this
public static void main(String[] args) {
String string= "11011100010000010001000000000000";
String string1= "00000000010000110000100000101100";
System.out.println(Integer.toHexString(Integer.parseInt(string1,2)));
System.out.println(Integer.toHexString(Integer.parseInt(string,2)));
}
the first string convert just fine but the second one has an error of java.lang.NumberFormatException
dont know what the problem is

try this:
Long.toHexString(Long.parseLong(string,2))
(edited from parsLong to parseLong)

For what's worth, you can also use the BigInteger class :
String string = "11011100010000010001000000000000";
String string1 = "00000000010000110000100000101100";
System.out.println(new BigInteger(string1, 2).toString(16));
System.out.println(new BigInteger(string, 2).toString(16));

When the most significant bit of a 32-character binary number is set to 1, the resultant value exceeds the range of positive numbers supported by int, and can no longer be interpreted as a valid integer number. This causes the exception according to the documentation:
An exception of type NumberFormatException is thrown if any of the following situations occurs:
The first argument is null or is a string of length zero.
The radix is either smaller than Character.MIN_RADIX or larger than Character.MAX_RADIX.
Any character of the string is not a digit of the specified radix, except that the first character may be a minus sign '-' ('\u002D') provided that the string is longer than length 1.
The value represented by the string is not a value of type int. (emphasis is mine)
In order to enter this negative binary value, use - sign in front of your number, and convert the remaining bits to 2-s complement representation.
If you need numbers that are longer than 32 bits, or if you would like the value to continue being interpreted as a positive number, you would need to switch to the 64-bit integer data type.

You can use Long instead of Integer, (Long.parseLong and Long.toHexString methods).

If you want to parse to integer, the range should be
10000000000000000000000000000000 to 01111111111111111111111111111111