CODEWITHSUNDEEP
javamathfloating-accuracypow
scv
scv

Reputation: 520

Does Java's Math.pow round off the result?

Today I came across a peculiar behavior of Math.pow(). I am not able to 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() produces the exact result as long as the parameters passed to it are integers or longs provided there is no overflow (which is true in this case).

Upvotes: 4

Views: 1935

Answers (2)

Jigar Joshi
Jigar Joshi

Reputation: 240908

because it casts long to double 

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

outputs:

1.0E18

and

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

outputs:

1000000000000000000

Upvotes: 10

Wickoo
Wickoo

Reputation: 7345

You can use BigDecimal for such calculations and not lose precision in converting between types:

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

Upvotes: 0

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