Result Java Math.pow completes the result?

Question

Today I came across a peculiar behavior of Math.pow (). I can not understand the output of the following java code:

long N1 = 999999999999999999L;
System.out.println("N1 : " + N1);

long N2 = (long) Math.pow(N1, 1);
System.out.println("N2 : " + N2);

I get the following output:

N1 : 999999999999999999
N2 : 1000000000000000000

I always thought that Math.pow () gives the exact result if the parameters passed to it are integer or long, if there is no overflow (which is true in this case).

Answer 1

because he drops longindouble

System.out.println((double)999999999999999999L);

outputs:

1.0E18

and

System.out.println((long)(double)999999999999999999L);

outputs:

1000000000000000000

• Why is this?

Answer 2

You can use BigDecimalfor such calculations and not lose the accuracy of conversion between types:

BigDecimal b = new BigDecimal(999999999999999999L);
BigDecimal pow = b.pow(1);