Develop Reference

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);
^

Getting number format exception [duplicate]

This question already has answers here:
java.lang.NumberFormatException: For input string
(7 answers)
Closed 6 years ago.
I am doing simple conversion from string to int but getting number format exception :
I have use below java Program :
String cId = "7000000141";
int iCid = Integer.parseInt(cId);
System.out.println(iCid);
Getting below exception :
Exception in thread "main" java.lang.NumberFormatException: For input string: "7000000141"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:459)
at java.lang.Integer.parseInt(Integer.java:497)
Why I am getting the above exception?

That's because it's out of range of integer. The maximum allowed value for integer is 2147483647
In Java, following are the minimum and maximum values.
width minimum maximum
int: 32 bit -2 147 483 648 +2 147 483 647
long: 64 bit -9 223 372 036 854 775 808 +9 223 372 036 854 775 807
source
Use a 'long' datatype instead
public class HelloWorld
{
public static void main(String[] args)
{
String cId = "7000000141";
long iCid = Long.parseLong(cId);
System.out.println(iCid);
}
}

Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: "11101110110100011110111011010001"

I wanna convert String to Binary Integer like this int k = 0B1101.....0111
and the error showen is :
Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: "11101110110100011110111011010001"
String a = "1110111011010001",
b = "11101110110101010110111011010001";
int K = Integer.parseInt(a.trim(),2);
int T = Integer.parseInt(b.trim(),2);

You get an exception because the 32-bit number that you pass represents a negative integer, so from parseInt's perspective it overflows an int.
You have two solutions to parse the number:
(1) Pass a negative representation with a minus sign, i.e.
System.out.println(Integer.parseInt("-0010001001011100001000100101111", 2));
or (2) pass the original number to parseLong, and cast the result to int, i.e.
System.out.println((int)Long.parseLong("11101110110100011110111011010001", 2));
Both snippets produce a negative value of -288231727 (demo).

This is because its above the Integer Limits of java.
Try using long instead of int.

Java - NumberFormatException. Convert long bynary string to dec

I want to convert binary string to dec.
public class HelloWorld{
public static void main(String []args){
System.out.println(Integer.parseInt("000011011111110111000001110000111110", 2));
}
}
I get error:
java.lang.NumberFormatException: For input string: "000011011111110111000001110000111110".
How to fix it?

Short solution - Integers simply don't go that high. That's not an int.
ParseInt() documentation mentions, you recieve a string and a radix, and get back the result of the convertion. However, integers are 4 bytes = 32 bits, and thus range from -(2^31) to 2^31-1, and your number - 11011111110111000001110000111110, is in fact 32 bits long - which means, it's bigger than the maximal value. Thus, the function throws this NumberFormatException - this is not a valid value for an int.
If you'd like to fix it, I'd use a ByteBuffer, like described here:
ByteBuffer buffer = ByteBuffer.wrap(myArray);
buffer.order(ByteOrder.LITTLE_ENDIAN); // if you want little-endian
int result = buffer.getShort(); // use with a bigInteger instead. you could use any of the bytebuffer functions described in the link :)

You can use BigInteger class and store the number as long:
BigInteger bigInt=new BigInteger("000011011111110111000001110000111110");
long a=bigInt.longValue();
The value you are going to store is too big for int and does not fall inside the range the type int can hold(-(2^31) to 2^31-1).So it throws NumberFormatException.long is a suitable option here.

You can use Long.parseLong for the string in your question, still you may find a limit in that also, so you have to implement different logic.
You can have a method that convert the binary string to integer.
public static long binaryToInteger(String binaryString) {
char[] chars = binaryString.toCharArray();
long resultInt = 0;
int placeHolder = 0;
for (int i=chars.length-1; i>=0; i--) {
if (chars[i]=='1') {
resultInt += Math.pow(2,placeHolder);
}
placeHolder++;
}
return resultInt;
}