This question already has answers here:
"Integer number too large" error message for 600851475143
(8 answers)
Java long number too large error?
(2 answers)
Closed 10 years ago.
I am writing a code in which I am trying to assign a value in long variable.
But java compiler is showing error that too large integer number.
I am trying to store 600851475143 in long type still.
class Sum {
static public void main(String args[]){
long num=600851475143;
}
}
append 'L' or 'l' at the end of the number to make it a long literal.you can use both lowercase(l)or uppercase(L), but uppercase(L) is recommend for readability.
long num=600851475143L;
An integer literal is of type long if it ends with the letter L or l;
otherwise it is of type int. It is recommended that you use the upper
case letter L because the lower case letter l is hard to distinguish
from the digit 1.
Reference
So use this -
long num=600851475143l;
or better
long num=600851475143L;
Related
This question already has answers here:
Integer to two digits hex in Java
(7 answers)
Closed 1 year ago.
I'm trying to pad a hexadecimal digit with 0's in the beginning so that the length of it is always 4. For example, the FF is going to be padded as 00FF. I have tried to use String.format("%04d", number) but it didn't work as my hexadecimal digit is actually a string.
I have also tried using StringUtils.leftPad() but for some reason, IntelliJ couldn't detect it.
Please note my count is going to change, but I haven't represented it here as I'm yet to implement it. This is my sample code. Thanks.
public static void main(String[] args) {
int count = 0;
String orderID = Integer.toHexString(count).toUpperCase();
System.out.println(String.format("%04d", Integer.valueOf(orderID)));
}
String.format("%04x", 255)
with output
00ff
(or use X for uppercases hex digits)
As #user16320675 point, more info about Formatter here.
This question already has answers here:
Alphanumeric increment algorithm in JAVA [closed]
(2 answers)
Closed 2 years ago.
I want to implement a functionality that auto increment an alpha numeric id in a Java program.
The ids types are Strings, have only digits and lower case latin letters, and is case insensitive.
Basically, I want a function called static String next(String id) that gives what should be the next id.
The id should increase from right to left, and new characters should be added on the left.
For example
assertEquals("1", IdUtils.next("0"));
assertEquals("a", IdUtils.next("9"));
assertEquals("10", IdUtils.next("z"));
assertEquals("10", IdUtils.next("0z"));
assertEquals("100", IdUtils.next("zz"));
assertEquals("zz1", IdUtils.next("zz0"));
assertEquals("zza", IdUtils.next("zz9"));
Thanks in advance.
A number with digits and lower case latin letters is a base-36 number, which is an extension of base-16, also known as hex.
So to increment the base-36 number, parse it, increment it, and convert back to text.
static String next(String id) {
return Long.toString(Math.incrementExact(Long.parseLong(id, 36)), 36);
}
Test
System.out.println(next("0"));
System.out.println(next("9"));
System.out.println(next("z"));
System.out.println(next("zzzzzz"));
System.out.println(next("dif5613ug"));
System.out.println(next("1y2p0ij32e8e6"));
System.out.println(next("1y2p0ij32e8e7")); // Long.MAX_VALUE, so will overflow
Output
1
a
10
1000000
dif5613uh
1y2p0ij32e8e7
Exception in thread "main" java.lang.ArithmeticException: long overflow
If you're going to have an "current ID" variable that you increment, then it's better to keep that state in an AtomicLong instead of a String. It's faster, and much easier to make thread-safe:
...
AtomicLong m_nextId = new AtomicLong();
static String next() {
long idnum = m_nextId.getAndIncrement();
return Long.toString(idnum,36);
}
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Closed 3 years ago.
when i want to get the result of 100/100000..
i got only 0.
Example
int one = 100;
int two = 100000;
int result = one/two;
toast(result); //Result is 0
Hey there "int" data type only stores integer values and not the decimals.
So if you divide 3 with 2 you would get 1 as answer instead of 1.5 .
Int just ignores the decimals .
You need to choose float or double data type for this to work.
Your variable named result must be declared and casted to float data type.
Appreciate the effort and mark this as answer if it helps you.....
This question already has answers here:
Initialize a long in Java
(4 answers)
The literal xyz of type int is out of range
(5 answers)
Why, In Java arithmetic, overflow or underflow will never throw an Exception?
(4 answers)
Closed 4 years ago.
Here are my 2 examples:
1. Directly assigned value to long data type:
long a = 12243221112432;
I get an error:
integer number too large
But when assisgned like this:
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(12);
al.add(24);
al.add(32);
al.add(21);
al.add(11);
al.add(24);
al.add(32);
long a=0;
for(int i=0; i<al.size(); i++){
a = a*100 + al.get(i);
}
System.out.println(a);
I get this output:
12243221112432
Why doesn't java throw an error in second example?
It isn't allowing to assign large value directly(example 1) but indirectly(example 2) it stores it and also allows me to use it too!
What is the reason for this to occur?
Is it because i am using integer in arraylist or something else?
UPDATE
Why is the large value stored in 'long a' in second example without using literal L?
It should have given me an error during 5th or 6th iteration of for loop...
Note
My question is not regarding the first example... I am asking why it worked for the second example...
Dont mark the question duplicate, since the other questions do not have my answer..stated above
For your assignment to work, you need to write it that way
long a = 12243221112432L;
This will indicate the compiler that your input is a long, not an int
integer number too large
as the error said, the number 12243221112432 is too large for an integer, it does not says that this number cannot fit into a long
Te be able to have this number, you have to make it a long by using the l or L indice : 12243221112432L
long : l L : 1716L
float : f F : 3.69F
double : d D : 6936D
This question already has answers here:
How does a leading zero change a numeric literal in Java?
(3 answers)
Closed 7 years ago.
How does this snippet of code prints "-511" as the output on the console?
class Test
{
public static void main(String[] args) {
int i = -0777;
System.out.printf("%d",i);
}
}
Is it to do with the way Java stores Negative numbers?
Integer numbers prefixed with 0 are octal numbers. To use decimal numbers remove the 0 prefix:
int i = -777;
Numbers starting with 0 are treated as being in octal by Java. -077 is equivalent to -63, which is what I get when I run your program.
When a number in java code starts with a 0, it interprets it as octal format