CODEWITHSUNDEEP
javaceil
sergiy
sergiy

Reputation: 161

Why isn't math.ceiling rounding up my number? java

Trying to write this script that works with inputs that often go up into hundreds of billions. However, Math.ceil won't round upwards? 

int clayturns  = (int)Math.ceil(clayneeded / 7500000000L);
System.out.println((int) Math.ceil(clayneeded / 7500000000L));
System.out.println("we need " + clayturns + "turns");
System.out.println("we need " + clayneeded + " clay added to our specific village.");

under here I have the eclipse console output. As you can see clayneeded isn't blank. As if that's the case then clayturns should always be something, it can't be zero, but it is.

we need 0turns
we need 3021588634 clay needed to our specific village.

Please let me know what you think. Does math ceil not work with Longs?

Upvotes: 1

Views: 744

Answers (3)

alt440
alt440

Reputation: 51

This is because you store your clayturns variable inside an integer. There is an Integer overflow happening, which returns 0.

The maximum value for an integer is +2,147,483,647.

Upvotes: -1

Jean-Baptiste Yunès
Jean-Baptiste Yunès

Reputation: 36401

Looks like clayneeded has integer type (probably long). So clayneeded/7500000000L is integer division (quotient), thus ceil computes the ceiling of an integer, so itself.

So you can compute it like this (if the remainder of the integer division is not nul then the ceiling must produce the next integer) :

long remainder = clayneeded % 7500000000L;
long result = (clayneeded/7500000000L)+(remainder!=0?1:0);

Upvotes: 0

Joe
Joe

Reputation: 607

is clayneeded an int/long? if you're doing int/long division, it'll truncate any fractional component before it gets to the Math.ceil. Given the size of your inputs, I'd stay away from floats/doubles, and use BigDecimals to do the math accurately here.

Upvotes: 3

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