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K&B Math.round() question

Ramesh Mangam
Greenhorn

Joined: May 03, 2005
Posts: 13
Hi All,

In K&B book, page 371 following definition is given for Math.round():

The round() method returns the integer closest to the argument. The algorithm is to add 0.5 to the argument and truncate to the nearest integer equivalent.


Going by the above definition:
round(8.2) = Nearest integer to (8.2 + 0.5 = 8.7) = 8
round(8.7) = Nearest integer to (8.7 + 0.5 = 9.2) = 9
round(-8.2) = Nearest integer to (-8.2 + 0.5 = -7.7) = -7 //?? (Ans: -8)
round(-8.7) = Nearest integer to (-8.7 + 0.5 = -8.2) = -8 //?? (Ans: -9)

But, for negative numbers, if I subtract 0.5 instead of adding, we get the right answers. So, I think it should be add 0.5 to +ve numbers and subtract 0.5 from negative numbers. Also, I don't think 'nearest integer' are the right words to use. I think it should be add 0.5 to +ve numbers or subtract 0.5 from negative numbers and 'discard decimal portion'. Does anyone have a different opinion on this?

Ramesh
shetal bansal
Ranch Hand

Joined: May 09, 2005
Posts: 63
I think that we have to add 0.5 to the given number and take its floor value, regardless of whether the number is positive or negative.
e.g. Math.round(-8.2) = Math.floor(-8.2+0.5)=Math.floor(-7.7)=-8
Sangita Mishra
Greenhorn

Joined: Jun 15, 2005
Posts: 22
yes, that's right. But just be careful as floor retuns a double but round returns an int or long.
narendra darlanka
Ranch Hand

Joined: Jun 17, 2005
Posts: 66
mathmetically integer closest to -7.7 in -8 not -7.so the algorithm is quite right
~naren


~naren<br /> scjp1.4<br /> scwcd1.4
narendra darlanka
Ranch Hand

Joined: Jun 17, 2005
Posts: 66
according to Khalid mughal its equivalent to adding .5 and taking floor of the result
~naren
Ramesh Mangam
Greenhorn

Joined: May 03, 2005
Posts: 13
mathmetically integer closest to -7.7 in -8 not -7.so the algorithm is quite right


Then how come the following is NOT working out:
round(8.2) = Nearest integer to (8.2 + 0.5 = 8.7) = 8

Nearest integer to 8.7 is 9. But, the correct answer is 8.
fred rosenberger
lowercase baba
Bartender

Joined: Oct 02, 2003
Posts: 10915
    
  12

I don't know about what the book says, but the correct method is to add 0.5, then take the floor - i.e. find the first integer smaller than whatever value you have.

so:

round(8.2) = floor of (8.2 + 0.5) = floor of (8.7) = 8
round(8.7) = floor of (8.7 + 0.5) = floor of (9.2) = 9
round(-8.2) = floor of (-8.2 + 0.5) = floor of (-7.7) = -8 (-7 is bigger)
round(-8.7) = floor of (-8.7 + 0.5) = floor of (-8.2) = -9 (-8 is bigger)

[corrected cut'n'paste goof]
[ July 01, 2005: Message edited by: fred rosenberger ]

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors
Jim Yingst
Wanderer
Sheriff

Joined: Jan 30, 2000
Posts: 18671
Fred, I think what you meant was more like

round(8.2) = Nearest integer to (8.2) = floor (8.2 + 0.5) = floor(8.7) = 8

We're not looking for the nearest integer to 8.7 - were looking for the floor of 8.7.

Similar changes can be made to the other 3 examples.
[ July 01, 2005: Message edited by: Jim Yingst ]

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fred rosenberger
lowercase baba
Bartender

Joined: Oct 02, 2003
Posts: 10915
    
  12

ahh... crud. that's what i get for a quick cut'n'paste. i'll correct it.
Fes D Gaur
Ranch Hand

Joined: Apr 29, 2005
Posts: 54
The K&B book also says that when if a number's fraction is less than 0.5 then round will work like Math.floot(); When the number's fraction is equal to or greater than 0.5 then round will work like Math.ceil().

When dealing with negative numbers think of them like this:

smaller number.........................................greater number
... -9.0, -8.9, -8.8, -8.7, -8.6, -8.5, -8.4, -8.3, -8.2, -8.1, -8.0,....-7.0,...

So if round(-8.2), since it is greater than -8.5 round is like the ceil() method and goes for the greater number which in this case is -8.0

Hope this helps,
Fes
 
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