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
Related
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));
I have this line of Java code that will throw NumberFormatException if the number represented as a String is above 2,147,483,647.
Because:
The int data type is a 32-bit signed two's complement integer. It has
a minimum value of -2,147,483,648 and a maximum value of 2,147,483,647
Code Throwing the NumberFormatException:
String largeNumberAsAString = "9999999999";
Integer.toHexString(Integer.parseInt(largeNumberAsAString)); // NumberFormatException
How can I get the same functionality of theInteger.toHexString() with a String parameter and not an int parameter due to NumberFormatException?
Use BigInteger to avoid numeric limits of primitive int and long:
BigInteger x = new BigInteger("9999999999999999999999"); // Default radix is 10
String x16 = x.toString(16); // Radix 16 indicates hex
System.out.println(x16);
The class conveniently exposes a constructor that takes a String, which gets interpreted as a decimal representation of a number.
Demo.
If your input value can be arbitrarily large, then #dasblinkenlight's answer involving BigInteger is your best bet.
However, if your value is less than 263, then you can just use Long instead of Integer:
String dec = "9999999999";
String hex = Long.toHexString(Long.parseLong(dec));
System.out.println(hex); // 2540be3ff
Live demo.
Use Integer.parseUnsignedInt
When the number is above 2^31 but below 2^32, thus in the negative int range,
you can do:
int n = Integer.parseUnsignedInt("CAFEBABE", 16);
(I used hexadecimal here, as it is easier to see that above we are just in that range.)
However 9_999_999_999 is above the unsigned int range too.
Try this way:
String largeNumberAsAString = "9999999999";
System.out.println(Integer.toHexString(BigDecimal.valueOf(Double.valueOf(largeNumberAsAString)).intValue()));
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.
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 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