足够大的双精度值返回Java错误

Question

I was basically playing with data types in order to learn java and here is what i am confused about.

double douVar = 30000000000.323262342134353;
int intVar = (int)douVar; // casting
System.out.println(intVar); 

Now converting the douVar to string i overwrite the double variable:

douVar = 345839052304598; // returns an error: Integer number too long
System.out.println(Double.toString(douVar));

complete error:

Error:(20, 18) java: integer number too large: 345839052304598

I'm using IntellijIDEA compiler. I didnt try this over eclipse though but i assume the compiler would not be so different. Does this mean that casting would result in modifying the original variable as well?

Answer 1

In Java, constant numbers without a decimal are ints by default. There are 2 ways to change the constant definition to a double:

1. Add a d or D at the end.
2. Add a .0 at the end

douVar = 345839052304598d;
douVar = 345839052304598.0;

Answer 2

Java interprets 345839052304598 as an integer.

In Java an Integer is 4 bytes long and therefore has a value range from –2 147 483 648 to 2 147 483 647 . (your number is obviously not in that range)

To make Java treat the number as a double, you must write:

double d = 345839052304598D;
/* or */
double d1 = 345839052304598.0;

(Same goes for a long , you have to write L instead of D in that case)

Answer 3

The literal 345839052304598 will be considered as an integer by the compiler , since this value is to large for an integer, a compile time error is thrown. Therefore you need to tell the compiler that the value you want to use is actually double value, that you can do the following way:

douVar =345839052304598D; // note D at the end which tells comipler that it is double

Answer 4

When you declare number as 345839052304598 compiler tries to read it as a 32 bit integer. But since this exceed the range of int compiler throws the error. You can use d, D or L suffix or add decimal place like .0 to to handle this,

Answer 5

Just change douVar = 345839052304598; to douVar = 345839052304598.00;