Why does this code throw NumberFormatException?
int a = Integer.parseInt("1111111111111111111111111111111111111111");
How to get the value of int for that String?
The value that you're attempting to parse is much bigger than the biggest allowable int value (Integer.MAX_VALUE, or 2147483647), so a NumberFormatException is thrown. It is bigger than the biggest allowable long also (Long.MAX_VALUE, or 9223372036854775807L), so you'll need a BigInteger to store that value.
BigInteger veryBig = new BigInteger("1111111111111111111111111111111111111111");
From BigInteger Javadocs:
Immutable arbitrary-precision integers.
This is because the number string is pretty large for an int . Probably this requires a BigInteger .
There is no integer value for that string. That's why it's throwing an exception. The maximum value for an integer is 2147483647, and your value clearly exceeds that.
Related
I'm trying to round a String value using Math.round as so:
String numberString = "61867345124";
int value = Math.round(Float.parseFloat(numberString));
but I'm getting value = 2147483647, what is going on here?
Actually, you have two issues here,
First, the output is 2147483647 because that is the max value an integer can store in Java. And your number is larger than that. You can confirm the max value for a primitive type by calling MAX_VALUE against its wrapper class, like below,
System.out.println(Integer.MAX_VALUE);
If the return value of Math.round is larger than Integer.MAX_VALUE it will return an integer having maximum value it can store,
From the documentation here,
If the argument is positive infinity or any value greater than or
equal to the value of Integer.MAX_VALUE, the result is equal to the
value of Integer.MAX_VALUE.
To resolve that you can store the result in a long typed variable.
Second, that will still not get you desired results because Math.round if provided with a float input will return an integer. You need it to return a long instead which is returned if you provide it a double value.
Here are the definitions, from the documentation of Math class.
static int round(float a)
static long round(double a)
To sum up you need to both store the output as a long as well as you need to parse the string value as double, like below,
String numberString = "61867345124";
long value = Math.round(Double.parseDouble(numberString));
System.out.println(value);
Looks like you are using a number way larger than an integer can contain, so the output is defaulting to the maximum integer value of 2147483647. You probably want to use a long to store that value.
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.
It's seems like integer but it not parse with Integer.parse() How can I fix it ?
int a= Integer.parseInt(lhs.getViewCount());
Caused by: java.lang.NumberFormatException: Invalid int: "6125635424"
at java.lang.Integer.invalidInt(Integer.java:138)
at java.lang.Integer.parse(Integer.java:378)
at java.lang.Integer.parseInt(Integer.java:366)
at java.lang.Integer.parseInt(Integer.java:332)
at com.youtubetracker.adapters.MyChannelListAdapter$1.compare(MyChannelListAdapter.java:46)
at com.youtubetracker.adapters.MyChannelListAdapter$1.compare(MyChannelListAdapter.java:37)
Always use BigInteger over long and float.
BigInteger value= new BigInteger(lhs.getViewCount());
You can convert that value to int , long or float.
Form Java Doc intValue()
Converts this BigInteger to an int. This conversion is analogous to a narrowing primitive conversion from long to int as defined in section 5.1.3 of The Java™ Language Specification: if this BigInteger is too big to fit in an int, only the low-order 32 bits are returned. Note that this conversion can lose information about the overall magnitude of the BigInteger value as well as return a result with the opposite sign.
The maximum Range of Integer is 2147483647 and you are converting 6125635424 to integer. This is the main reason behind the Exception.
Long.parseLong(lhs.getViewCount());
Will solve your problem.
Try this :
Replace: int a= Integer.parseInt(lhs.getViewCount());
With
Long a= Long.parseInt(String.valueOf(lhs.getViewCount()));
'6125635424' cannot store in Integer you need to store in Double or Long.
Try this
Long a = Long.parseLong(lhs.getViewCount().toString());
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);