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I have a question about an error when translating the pseudo code below into Java. There is no problem with the loss of accuracy because of truncation but a programming language issue that I can't seem to find.

I though the answer would be that the result of Math.pow() would be too much for a double (range from 4.94065645841246544e-324d to 1.79769313486231570e+308d. But apparently there is more wrong and I can't see it.

PSEUDO CODE

x = 7.0
y = x^1000.0
n = y truncated to an integer

converted to

JAVA

double x, y;
int n; 

x = 7.0;
y = Math.pow(x,1000.0);
n = (int)y;

Much appreciated, I'd really like to understand the issues here.

JasonMArcher
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Matt Russell
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    What error are you talking about? There's nothing wrong with the code you've posted. – azurefrog Nov 17 '14 at 05:40
  • What result do you get and what do you expect? – Elliott Frisch Nov 17 '14 at 05:40
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    7^1000 is about 1.25e845. It exceeds the range of double. long, 64bit integer, can express -9.2e18 to +9.2e18. 1.25e845 exceeds the range of long also. This is problem of language limit of java, not the translation. – Fumu 7 Nov 17 '14 at 05:50

1 Answers1

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When you do,

double x = 7.0;
double y = Math.pow(x, 1000.0); // <-- INFINITY

you do exceed the precision of double and you get Double.POSITIVE_INFINITY

System.out.println(y);

Output is

Infinity

When you cast that to int you get Integer.MAX_VALUE which is 2^31-1 or 2147483647. If you want to get a more convincing result you could use BigInteger like

int n = BigInteger.valueOf(7).pow(1000).intValue();
System.out.println(n);

Output is

432069569

Of course, the real result is obtained by

System.out.println(BigInteger.valueOf(7).pow(1000));

Which results in a very big number indeed.

Elliott Frisch
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  • So, it's both the fact that y = infinity and you get the max integer value for n when casting. I didn't know the second occurs. Thanks so much. – Matt Russell Nov 17 '14 at 06:08