sqrt requires int but I'm already using int

Question

int Asum,temp4;
temp4=Math.sqrt(Asum);

This is part of a code I'm writing. I've declared Asum and temp4 as integers. I'm trying to get the sqrt of Asum and assign it to temp4. but for some reason, java gives me the error : Possible loss of precision. reqd=int. found= double. I need temp4 and Asum to be an integer. and it's not necessary to get the precise decimal values, I just need the rounded off integers.

Read the javadoc of Math.sqrt(). What's the return type? Is it int? — JB Nizet, Apr 19 '15 at 17:03

score 1 · Accepted Answer · answered Apr 19 '15 at 17:03

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Have you tried to add an explicit cast to int so that the compiler knows that you know what you are doing?

temp4= (int) Math.sqrt(Asum);

answered Apr 19 '15 at 17:03

FooTheBar

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score 1 · Answer 2 · answered Apr 19 '15 at 17:03

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temp4 shouldn't be integer... it could be float.

answered Apr 19 '15 at 17:03

tanaydin

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It **must** be `double`, else OP has to cast the result to the appropriate type. – Svetlin Zarev Apr 19 '15 at 17:07

score 0 · Answer 3 · answered Apr 19 '15 at 17:06

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I think you need to make use of the Math.round function.

Try:

temp4=Math.round(Math.sqrt(Asum));

answered Apr 19 '15 at 17:06

Tech Lobster

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it still gives me the same error, but it'll do for my purposes. thanks! – SirVirgin Apr 19 '15 at 17:11

sqrt requires int but I'm already using int

3 Answers3