This question already has answers here:
hex to int number format exception in java
(4 answers)
Closed 6 years ago.
I get following exception:
java.lang.NumberFormatException: For input string: "3693019260"
while calling
Integer.parseInt(s);
And I don't know why I get it.
3693019260 is smaller than 2^32
3693019260 is clearly a natural number
the String is cleared from all non-digit elements with s.replaceAll("[^0-9]", "")
So why do I get this exception?
According to the little bit of debugging I did, I saw that the number dips under multmin but I don't know what this variable does and how I should interpret this observation.
While 3693019260 may fit into a 32 bit unsigned integer, it looks like you are trying to parse it into a plain int which is a signed integer. Signed simply means that it supports negative values using -.
With signed numbers, half of the namespace is reserved, so your number must fit into 2^32÷2−1 2147483647 instead of simply 2^32.
The simplest fix is to parse the value as a long instead of an int. Long numbers are 64 bits and support many more digits in the string.
Related
This question already has answers here:
Are there binary literals in Java?
(2 answers)
In Java, can I define an integer constant in binary format?
(8 answers)
What are the Java primitive data type modifiers?
(3 answers)
Java 7 underscore in numeric literals
(8 answers)
Closed 2 years ago.
I'm practicing with some tasks of a Java course and I came across this variable
int x = 0b1000_1100_1010;
I know that a "f" and a "d" beside a number means that the number is a float or a double, respectively. But what about this "b" between the number?
I saw here that this is related to bytes, but I didn't quite understand how it works.
My question also applies to the "x" between numbers that I just saw on that link.
Thanks!
This is the binary literal.
It's a notation to show that the number is represented in binary.
It's like when you use the hexadecimal notation: 0xF9. In Java, you can use 0b1111_1001 to represent the same number, which is 249 in decimal.
It's not related to bytes, but to bits. You can clearly see which bits are set and which are not. By default a number starting with 0b is an int, but you can write a long like this 0b1010L (note the trailing L).
The b can be lowercase or uppercase. So this is also valid: 0B1111. Note that since the 0b prefix indicates a binary representation, you're not allowed to use any character other than 0 and 1 (and _ to mark a separation).
Java allows you to use _ in int values as shown below:
public class Main {
public static void main(String[] args) {
int x = 1_2_3;
System.out.println(x + 100);
}
}
Output:
223
The byte data type is an 8-bit signed two's complement integer. It has a minimum value of -128 and a maximum value of 127 (inclusive).
The byte data type can be useful for saving memory in large arrays,
where the memory savings actually matters. They can also be used in
place of int where their limits help to clarify your code; the fact
that a variable's range is limited can serve as a form of
documentation.
https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
This question already has answers here:
Converting 32-bit binary string with Integer.parseInt fails
(4 answers)
Closed 5 years ago.
The binary string is "10101010101010101010101010101010", 32 bits
Got this exception when i try
Integer.parseInt("10101010101010101010101010101010", 2);
But the same string(add "0b" prefix)
System.out.print(0b10101010101010101010101010101010);
return -1431655766.
Is this a valid binary string?
Integer.parseInt() is the wrong method as it only accepts signed numbers.
With Java8, use:
Integer.parseUnsignedInt("10101010101010101010101010101010", 2);
Before Java8, use:
(int) Long.parseLong("10101010101010101010101010101010", 2);
The value you wanted to convert goes beyond the int type size, UseBigInteger:
BigInteger b = new BigInteger("10101010101010101010101010101010",2);
Integer.parseInt("10101010101010101010101010101010", 2);
Note that this would be 2863311530, which would cause an overflow, as it is above Integer.MAX_VALUE.
System.out.print(0b10101010101010101010101010101010);
This however uses the internal representation of an integer, which is two-complement form. Thats why it is negative. The Integer.parseInt() however treats it as an unsigned binary number, which causes the Exception.
It is important to understand the different bit representations.
Edit: If you want your input to be interpreted as two-complement, use this:
Integer.parseUnsignedInt("10101010101010101010101010101010", 2);
This question already has answers here:
Why are integer literals with leading zeroes interpreted strangely?
(8 answers)
Closed 6 years ago.
I just wanted to experiment with some integers and assigned the value "0013" to an integer a. When I print the value to the output console I get "11". What causes this? Why I do not get 13 ?
int b = 0013;
System.out.println(b);
A leading zero mean octal. Just like a leading 0x mean hexadecimal
Java has accidentally automatically taken your number to be an octal number. Unlike hexadecimal notation, where the number is preceded by a 0x, octal is proceeded by a single zero. The compiler has probably taken your number and made it into an octal format.
Try using 13 instead of 0013
This question already has answers here:
Why doesn't Java support unsigned ints?
(17 answers)
Closed 6 years ago.
Byte value is between 0 and 255, very simple and straight forward. Java didn't think so however and decided that values to be between -128 and 128 and ruined my life and many others.
I just want to know what's the big idea? why everytime i need to get the unsigned byte value do I have to do this:
int byteValue = (int) javaByte & 0xFF;
Java supports only signed integers. byte is integer number as well so Java is just consistent here.
This is not case for C for example where char type doesn't have defined whether it's signed or unsigned. You need to explicitly tell. This is in my opinion much worse than saying that it's by default signed.
Anyway, the original intention in C for char was a string character where the unsigned made sense. In Java there is different type for string character - char.
This question already has answers here:
Assigning int to byte in java?
(5 answers)
Closed 8 years ago.
I have an int variable in java whose value is less than 256. So Can I store it in a Single byte variable.
I performed the following code.
int i =247;
byte b = (byte) i ;
But When I print it,
System.out.println(" i = "+(int)b);
It outputs
i = -9
Is it possible to convert int values less than 256 to single byte variables.
Pls. Help..
This isn't going to work. The MAX_VALUE for byte is 127 because bytes are signed in Java. (Consequently, the MIN_VALUE is -128. This is your typical Two's Compliment pattern.)
That being said, you have a couple options:
Leave it as an int and just use an int for your calculations. This might even be faster for you.
Leave it as a byte even if the number isn't correct. The numbers might not be correct in Java, but if all you care about is the bits being correct, they will be correct. So you can send that bite (which is a -9 in Java) across a webservice to another app that reads it as an unsigned byte, and it will read it as 247 correctly.