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I'm getting java.util.InputMismatchException when the input type matches the expected type [duplicate]

This question already has answers here:
Scanner for long integer, Exception in thread "main" java.util.InputMismatchException
(4 answers)
Closed 3 years ago.
This code is in the main method:
Scanner input = new Scanner(System.in);
int number = input.nextInt();
I need to input a 16-digit integer. I get java.util.InputMismatchException.
I cannot figure why this is, as an int is expected and I enter an int. Is 16 digits too long?

Integer goes from -2147483648 to 2147483647 so... yeah... 16 digits is too 'Long'.
Try long which max value is 9223372036854775807.

input string cannot be parsed to integer in java [duplicate]

This question already has answers here:
java.lang.NumberFormatException: For input string
(7 answers)
Closed 5 years ago.
i have following numbers saved in array (readed from XML files):
100000000000008261
100000000000008266
100000000000008267
100000000000008268
The SeqNrList is filled by this:
ArrayList SeqNrList = new ArrayList<>();
SeqNrList.add(doc.getElementsByTagName("SequenceNumber").item(0).getTextContent());
I've try to get the minimum and maximum value with following code:
int seqsizemin = Integer.parseInt((String) Collections.min(SeqNrList));
int seqsizemax = Integer.parseInt((String) Collections.max(SeqNrList));
Also, i've try'd with following:
int seqsizemin = Integer.valueOf((String) Collections.min(SeqNrList));
int seqsizemax = Integer.valueOf((String) Collections.max(SeqNrList));
But i got only following error when i run my script:
Exception in thread "main" java.lang.NumberFormatException: For input string: "100000000000008261"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:583)
at java.lang.Integer.parseInt(Integer.java:615)
at ReadXMLFile.main(ReadXMLFile.java:117)
Is there any special function needed, why i cant save

According to the JavaDoc of Integer#parseInt():
An exception of type NumberFormatException is thrown if any of the following situations occurs: [...] The value represented by the string is not a value of type int.
Any number which can not be parsed into an int is invalid.
In your case is the number 100000000000008261 larger than the 32-Bit Integer of Java. That's why you're getting the error.
To go arround this, have a look at Long#parseLong():
long seqsizemin = Long.parseLong((String) Collections.min(SeqNrList));
long seqsizemax = Long.parseLong((String) Collections.max(SeqNrList));

You need to remember that integer data types can hold value requiring up to 32 bits, the values from your example require more than 32 bits to be represented, using Long.parseLong could give you the value represented as long and in case you need to handle bigger values, take a look to BigInteger.

Storing Value as a Short

I am not sure why I keep getting this error but I was hoping someone could help me. I am trying to work on a partial MIPS disassembler program in Java but when trying to store my I format offset (16 Bits) I get an error message(Stars where the error occurs):
private short offset;
public void setOffset (String binary) {
**short s = Short.parseShort(this.binary.substring(15, 31),16);**
offset = s;
}
public short getOffset(){
return offset;
}
This is th error message I get:
Exception in thread "main" java.lang.NumberFormatException: For input string: "1000000000001100"
at java.lang.NumberFormatException.forInputString(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at java.lang.Short.parseShort(Unknown Source)
at HexI.setOffset(HexI.java:65)
at MipsDisassembler.main(MipsDisassembler.java:31)
Essentially I would like to store it as a short and use bitwise operators to return a hex value

short s = Short.parseShort(this.binary.substring(15, 31),16);
You are specifying radix 16 (hexadecimal) for the input when in actuality it should be 2 since the input is binary.
It should be
short s = Short.parseShort(this.binary.substring(15, 31),2);
^

parseInt throws error

I have this small piece of code in java which throws the following error when the code is run
Exception in thread "main" java.lang.NumberFormatException: For input string: "10000000000"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:495)
at java.lang.Integer.parseInt(Integer.java:527)
at hello.main(hello.java:6)
public class hello {
public static void main(String args[])
{
int x = 1024;
String h = Integer.toString(x, 2);
int xx = 9*(Integer.parseInt(h));
System.out.println(xx);
}
}
I suspect that this problem is related to the size of the values/parseInt. Can you please explain the reason for this error to me in detail.

This is because this surpasses the maximum value for an integer of 2,147,483,647

You are getting java.lang.NumberFormatException: For input string: "10000000000" because it exceed the range of int.
integer is a signed 32-bit type that has a range from –2,147,483,648 to 2,147,483,647.
long is a signed 64-bit type and is useful for those occasions where an int type is not large enough to hold the desired value, range is from –9,223,372,036,854,775,808 to
9 ,223,372,036,854,775,807. This makes it useful when big, whole numbers are needed.
Try this line of code-
long xx = 9*(Long.parseLong(h));

You are receiving this error because you are trying to parse a value too large for the Integer type. Try using Long instead.

Cannot convert String to Integer in Java [duplicate]

This question already has answers here:
How do I convert a String to an int in Java?
(47 answers)
Closed 5 years ago.
I have written a function to convert string to integer
if ( data != null )
{
int theValue = Integer.parseInt( data.trim(), 16 );
return theValue;
}
else
return null;
I have a string which is 6042076399 and it gave me errors:
Exception in thread "main" java.lang.NumberFormatException: For input string: "6042076399"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:461)
Is this not the correct way to convert string to integer?

Here's the way I prefer to do it:
Edit (08/04/2015):
As noted in the comment below, this is actually better done like this:
String numStr = "123";
int num = Integer.parseInt(numStr);

An Integer can't hold that value. 6042076399 (413424640921 in decimal) is greater than 2147483647, the maximum an integer can hold.
Try using Long.parseLong.

That's the correct method, but your value is larger than the maximum size of an int.
The maximum size an int can hold is 231 - 1, or 2,147,483,647. Your value is 6,042,076,399. You should look at storing it as a long if you want a primitive type. The maximum value of a long is significantly larger - 263 - 1. Another option might be BigInteger.

That string is greater than Integer.MAX_VALUE. You can't parse something that is out of range of integers. (they go up to 2^31-1, I believe).

In addition to what the others answered, if you have a string of more than 8 hexadecimal digits (but up to 16 hexadecimal digits), you could convert it to a long using Long.parseLong() instead of to an int using Integer.parseInt().