Code:
String myVar = "1255763710960";
int myTempVar=0;
try
{
myTempVar = Integer.valueOf(myVar);
}
catch (NumberFormatException nfe)
{
System.out.println(nfe.toString());
}
Output:
java.lang.NumberFormatException:
For input string: "1255763710960"
I have absolutely no idea why this is.
The value you're trying to store is too big to fit in an integer. The maximum value for an Integer is 231-1, or about 2 billion. This number exceeds that by several orders of magnitude.
Try using a Long and parseLong() instead.
Java Integer maximun value is 2^31-1=2147483647
You should use Long.valueof()
Your String representation is too big (>Integer.MAX_VALUE) for parsing to an int. Try a long instead.
1255763710960 is more than Integer.MAX_VALUE which is 2147483647, so that value doesn't fit in an int.
You'll need to use a long and Long.valueOf() (or better yet Long.parseLong() to avoid unnecessary auto-unboxing) to parse that value.
Related
I have the below json property which holds the numeric value
"creditLimit": "844500"
which has the following conditions:
Should not exceed 10 digits
Must be a whole number, should not have decimals
Below are example valid inputs for which I want to throw a validation error message back to the user, saying invalid entry:
45500.00
9876543210
Example invalid inputs:
540.50
98765432109
When I tried
Double.valueOf(ent.creditLimit).intValue()
the final value alters, looks like it round off the value.
I do not want to keep the decimals.
How to retain the exact value? Thanks in advance
You could either use a regular expression to validate or you could parse as a BigDecimal then use intValueExact():
BigDecimal bd = new BigDecimal("540.50");
try
{
bd.intValueExact();
//number is a whole number
} catch(ArithmeticException e)
{
//number is not a whole number
}
Ensure there is a javax.validation implementation in your classpath/runtime.
In your JSON-Pojo, then:
#javax.validation.constraints.Pattern(regexp = "^\\d{1,10}$")
private String creditLimit;
Allows only numerical strings of length 1 to 10. Throws exception otherwise.
I get this error whenever I run my program:
Exception in thread "main" java.lang.NumberFormatException: For input string: "9999997560"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:461)
at java.lang.Integer.valueOf(Integer.java:554)
at jsobcpuburnerpd3.Main.main(Main.java:22)
Java Result: 1
The program has two do-while loops that are nearly identical, and the first works perfectly. I only have problems with the second one.
BigDecimal lower = new BigDecimal("1000001520");
BigDecimal upper = new BigDecimal("9999997560");
int var = 2520;
String strL = lower.toString();
Integer intL = Integer.valueOf(strL);
String strU = upper.toString();
Integer intU = Integer.valueOf(strU);
Both numbers have the same amount of digits, and are converted to Integer the same way. Both are handled nearly the same way in the loops.
intL = intL + var;
intU = intU - var;
I have also attempted this without converting from BigDecimal to String, and inputting the number directly to String, but still got the same error.
EDIT I am using BigDecimal because it's a part of what my teacher wanted us to use in our work. I know it's not needed.
Obviously you will get a number format exception because the maximum value that can be store in an Integer is 2,147,483,647 which is less than 9,999,997,560.
On the other hand 1,000,001,520 is less than 2,147,483,647 which is why that works fine.
(Edit Based on Tunaki's suggestion) FYI - Moreover you really don't have to use a BigDecimal because it seems that the only reason you need it is to convert it to an Integer. So it is not required.
Also an Integer is not required because you don't seem to be requiring a reference type and hence the primitive type int should be apt. Moreover in your for loop you are adding and subtracting values from Integer which will lead to unnecessary boxing and unboxing and will be very inefficient. So use the primitive type int which would be better.
Convert the BigInteger to Integer using intValue() Method
BigDecimal lower = new BigDecimal("1000001520");
BigDecimal upper = new BigDecimal("9999997560");
Integer intBL = lower.intValue();
Integer intBU = upper.intValue();
My issue is that
checksum = Long.parseLong("-986.9");
is giving NumberFormatException.
Isn't this a parsible long?
A Long is not a decimal. Use Double and convert :
Double.parseDouble("-986.9").longValue();
No it isn't. A long is a numeric integer type, and your number has a decimal point.
You want a double here (ie Double.parseDouble()).
long is an integer data type, -986.9 is not an integer. The below works for me.
long checksum = Long.parseLong("-986");
System.out.println(checksum);
double checksum2 = Double.parseDouble("-986.6");
System.out.println(checksum2);
Use a double, use the method Double.parseDouble(var)
In my java code, I have a string 9632580147, and when I convert it into a int, using this code:
try{
sNumberInt = Integer.parseInt(sNumber);
} catch(NumberFormatException nfe) {
Log.d("NUMBER", nfe.getMessage());
return;
}
It goes into the catch block saying Invalid int: "9632580147"...
Does anyone know how to fix this?
Thanks
When you type
int sNumber = 9632580147;
into your code, the compiler will tell you:
The literal 9632580147 of type int is out of range
The reason is that your number is too big to fit into an int, use a long instead.
Max value of int is 2147483647 and you are trying to pass 9632580147 which is greater. Try maybe Long.parseLong(sNumber)
It seems you are passing value larger than 32 bits:
9632580147 = 1000111110001001011000001000110011 (34 bits)
Integer max value is 2147483647. If you want to part that number, you need to parse it into a Long.
The maximum value of the integer in Java is 2147483647 while your input 9632580147 is greater. Instead, use a long data type:
long sNumberLong = Long.parseLong(sNumber);
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().