Java-TypeCasting Question

Question

long a = (long)Math.pow(2, 32);
// a = 4294967296 :)

long a = (int)(long)Math.pow(2, 32);
//a = 0 ?!

long a = (int)Math.pow(2, 32);
//a = 2147483647 WTF??!!!

The first expression is obvious. a is printed as is.

The second expression is a bit confusing. Great importance

10000000000000000000000000000000000 // 1, followed by 32 ZEROs, 33 bits in all

When it is forced into an int, how is it done as ZERO? Shouldn't he take the most important 1s as a sign bit and assume that the number is -2147483648? [REMOVAL OF THE CASE]

Also, when the double returned from Math.pow (4.294967296E9) is directly converted to int, why is it 2147483647?

I read the casting type and data types from the book, but the text does not explain much. I'm confused. Please explain why the second and third expressions give these results.

Answer 1

, 32 , , 1 (, , ).

4294967296 = 0x100000000 ( ), 32 .

: , ​​int, , , int, .

Answer 2

5.1.3 Java .

http://java.sun.com/docs/books/jls/second_edition/html/conversions.doc.html#25363

n , , . , , . .

int value = (int)(long)(Math.pow(2, 31)); // double to long to int
System.out.println(value); // Prints -2147483648

long lvalue = value; // int back to long
System.out.println(value); // Prints -2147483648 again

- , , , . /, float / , , .

Answer 3

spec, 5.1.3. T - , , char int.

Narrowing the conversion of a signed integer to an integral type T simply discards everything except n bits of the least significant bit, where n is the number of bits used to represent type T. In addition to the possible loss of information about the value, a numerical value can lead to the sign of the resulting values ​​will differ from the sign of the input value.

You fall into the lower digits, which makes this case equal to 0.