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java.lang.NumberFormatException.forInputString

I'm trying to convert my binary string to decimal value but i'm getting error.
The binary string is of 32 bits so no case of overflow.
String s = "11111111111111111111111111111101";
System.out.print(s.length());
System.out.print(Integer.parseInt(s,2));
Exception is:-
Exception in thread "main" java.lang.NumberFormatException: For input string: "11111111111111111111111111111101"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:583)

Integer type is not big as you expect, instead, try to use BigInteger:
new BigInteger(s, 2)
Output
324294967293

The first bit is 1, so this number cannot be represented as a positive int. If you try, it will "overflow" and become negative.
If that is what you want, you can parse the string as an unsigned value:
System.out.print(Integer.parseUnsignedInt(s,2));
Output: -3

You can use long:
String s = "11111111111111111111111111111101";
System.out.print(s.length());
System.out.print(Long.parseLong(s,2));

number format exception when trying to parse a string

I want to convert Integer.MAX_VALUE to binary and want the represenation to be of type int. I passed it to Integer.tobinarystring() and wraped that with Integer.parseint but i get numberformatexception.
Here is the code
System.out.println(
Integer.parseInt(
Integer.toBinaryString(Integer.MAX_VALUE)
)
);
Here is the exception
Exception in thread "main" java.lang.NumberFormatException: For input string: "1111111111111111111111111111111"

Integer.MAX_VALUE is 2,147,483,647
In binary this is:
1111111111111111111111111111111
If we treat that like an integer again, which is 1,111,111,111,111,111,111,111,111,111,111 you can probably see that it is much larger than the max value.
You probably want to look into BigInteger if you really need to deal with that as a int.

If you want to get the integer value of the binary string
1111111111111111111111111111111
you must use another signature of parseInt() that takes as 2nd parameter the radix, in this case of a binary string the radix is 2
String str = Integer.toBinaryString(Integer.MAX_VALUE);
int number = Integer.parseInt(str, 2);
System.out.println(number);
it will print:
2147483647

NumberFormatException when converting Integer String in Java

I declare this variable:
private String numCarteBancaireValide=String.valueOf(((Integer.parseInt(Config.NUM_CARTE_BANCAIRE_VALIDE) ) + (Integer.parseInt("0000000000000001"))));
Config.NUM_CARTE_BANCAIRE_VALIDE is a string.
After Execution, I receive this error message :
java.lang.NumberFormatException: For input string: "4111111111111111"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:583)
at java.lang.Integer.parseInt(Integer.java:615)
Please, Can you give your advices ?

Use Long.parseLong(), as your parameter is too large for an Integer. (Maximum for an integer is 2147483647)
PS: using Integer.parseInt("0000000000000001") doesn't make much sense either, you could replace this with 1.

The 4111111111111111 (which most likely is the value of Config.NUM_CARTE_BANCAIRE_VALIDE) overflows the Integer type.
Better try with:
//if you need a primitive
long value = Long.parseLong(Config.NUM_CARTE_BANCAIRE_VALIDE);
or
//if you need a wrapper
Long value = Long.valueOf(Config.NUM_CARTE_BANCAIRE_VALIDE);

The maximum value of integer is 2147483647. So you need to use Long.parseLong instead to parse 4111111111111111. Something like this:
long l = Long.parseLong("4111111111111111");
On a side note:
As Alex has commented, if this number is representing a credit card number then you can treat it like a string instead of changing to long as there is no arithmetic calculations involved with the credit card numbers.

Integer.parseInt will attempt to parse an integer from a String.
Your "4111111111111111" String does not represent an valid Java integer type, as its value would be > Integer.MAX_VALUE.
Use Long.parseLong instead.

Problems parsing String to long - what am I doing wrong?

Here is the code I am working with in a main method:
String numbers = "12345678900";
long upc = Integer.parseInt(numbers);
System.out.println(upc);
gives me:
Exception in thread "main" java.lang.NumberFormatException: For input string: "12345678900"
at java.lang.NumberFormatException.forInputString...
at java.lang.Integer.parseInt....
at java.lang.Integer.parseInt...
at testRun.main...
I cannot use a double, they need to be stored as values without a decimal. I am trying to get the string of numbers from a string into a variable that holds numbers (no decimals)

To parse a long, use Long.parseLong, not Integer.parseInt. That way, you get access to the full range of long values (whereas with parseInt, you only get the rather more restricted range of int values).

Use Long.parseLong()
String numbers = "12345678900";
long upc = Long.parseLong(numbers);
System.out.println(upc);

use
String numbers = "12345678900";
long upc = Long.parseLong(numbers);
System.out.println(upc);

The number you are passing is outside the range of integer which is from -2,147,483,648 to 2,147,483,647.
Try using the static function parseLong

Your number is enough long to not fit in Integer. So, you can't cast in integer. You can cast it in Long by calling Long.parseLong(number).

Use
long upc = Long.parseLong(numbers);

Converting from hex to int in java

I am getting an error because of the following line of code:
int x = color(Integer.parseInt("ffffffde",16));
I think it might be because it is a minus value
Any ideas why or how or how to fix it?
EDIT:
Sorry, didn't include the actual error. here it is:
Exception in
thread "Animation Thread" java.lang.NumberFormatException: For input
string: "ffffffde" at
java.lang.NumberFormatException.forInputString(Unknown Source) at
java.lang.Integer.parseInt(Unknown Source)
EDIT 2:
The value ("ffffffde") is being created by the following code:
Integer.toHexString(int_val);
EDIT 3:
Turns out it is a known bug (http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4215269)
Although you can convert integers to hex strings, you cannot convert them back if they are negative numbers!!

ffffffde is bigger than integer max value
Java int is 32 bit signed type ranges from –2,147,483,648 to 2,147,483,647.
ffffffde = 4,294,967,262
Edit
You used Integer.toHexString(int_val) to turn a int into a hex string. From the doc of that method:
Returns a string representation of the integer argument as an unsigned integer in base 16.
But int is a signed type.
USE
int value = new BigInteger("ffffffde", 16).intValue();
to get it back as a negative value.

If you are getting error like this,
Exception in thread "main" java.lang.NumberFormatException: For input string: "ffffffde"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:461)
at com.TestStatic.main(TestStatic.java:22)
Then there is problem with value you are passing that is ffffffde . This is not a valid hex value for parsing to int.
Please try this
int x = Integer.parseInt("ffffde",16);
System.out.println(x);
It should work.
For hex values more than that you have to pars to Long
Long x = Long.parseLong("ffffffde",16);
System.out.println(x);
And this also should work