This question already has answers here:
Datatype to store 20 digit number
(4 answers)
Closed 6 years ago.
I am trying to convert 10000000000000000000000000000000 to BigInteger.
But it is not converting.
My code is
BigInteger number = BigInteger.valueOf(10000000000000000000000000000000);
It is showing The literal 10000000000000000000000000000000 of type int is out of range.
Is there any way to convert this number as I have to use this number as integer in my programme?
The whole point of BigInteger is to support numbers long and int cannot so you can't use those. What you can do is use a String.
BigInteger bi = new BigInteger("10000000000000000000000000000000"):
Related
This question already has answers here:
Beginners Java Question (int, float)
(4 answers)
Division of integers in Java [duplicate]
(7 answers)
Closed 2 years ago.
If I divide a double type variable, the Decimal part becomes zero.
a=13122/10;
System.out.println (a);
Prints
1312.0
As you can see, the Decimal part became zero when I divided it.
But I need the value
1312.2
Your problem is that while you may have stored "a" as a double, you are really dividing two "ints" and saving that. When you divide 2 ints, the number automatically gets rounded down. So, it's rounded down to 1312.0.
What you need is this,
a = (double)13122/10;
or this:
a = 13122.0/10;
You are dividing integers. You can cast the division and then you are ok
double a = (double) 13122/10;
System.out.println (a);
This question already has answers here:
is there anyway to convert from Double to BigInteger?
(3 answers)
Closed 5 years ago.
How do you convert a double to a BigInteger, as accurately as possible? BigInteger doesn't have a valueOf(double) method. It does have a valueOf(long) method, but converting via long, would destroy any value outside the 2^64 range, and in particular I'm trying to get large numbers like 1e100 to convert to their corresponding integer values to the sixteen or so significant figures allowed by floating point precision.
You can convert the double into a BigDecimal and then into a BigInteger:
BigInteger k = BigDecimal.valueOf(doublevalue).toBigInteger();
You could use BigDecimal.valueOf(double) and then call toBigIntegerExact() on that. Like,
BigInteger bi = BigDecimal.valueOf(Math.PI).toBigIntegerExact();
This question already has answers here:
Convert integer into byte array (Java)
(11 answers)
Closed 8 years ago.
Suppose i have the number 520 that is mapped in two bytes that gives me the number: 1000001000 and i want to convert this number (520) to two other numbers, these numbers should be: 2 and 8 because 00000010 will give me 2 and 00001000 will give me 8. how can i do this with java?
Like this:
int theNumber = 520;
byte oneNumber = (byte)theNumber;
byte otherNumber = (byte)(theNumber >> 8);
This question already has answers here:
What is JavaScript's highest integer value that a number can go to without losing precision?
(21 answers)
Closed 8 years ago.
I have this in java:
int minimo = Integer.MAX_VALUE;
how would it look in javascript? Is there an equivalent of n"integer" in javascript?
var minimo = Integer.MAX_VALUE;
?
JavaScript has only one number type Number. It automatically switches between floats and integers as needed.
Use Number.MAX_SAFE_INTEGER to get the max integer value and Number.MAX_VALUE to get the max float value.
For me these are 9007199254740991 and 1.7976931348623157e+308
This question already has answers here:
Print an integer in binary format in Java
(24 answers)
Closed 8 years ago.
Say I have an integer number: 11728322732
how can i convert it to a string according to its byte representation, i.e 11100011000000001100000111110001 (32 bits \ 4 bytes \ integer)
Thanks
Here it is:
Long.toBinaryString(11728322732L);
Actually, 11728322732 is not an Integer but a Long (because it is greater than Integer.MAX_VALUE). So if you really want to convert this long to a 32-bits int (can't figure why actually), you could do:
Integer.toBinaryString((int)11728322732L);