I think banks do this rounding of .5 to the nearest even integer because this kind of value (.5) occurs quite a lot in monetary amounts. It has the effect of distributing evenly the rounding down and rounding up occurrences.

That can be one reason... But problem occurs if even-odd distribution has lot of variance say 100 7.5 and 1000 8.5 ......

This will be unlikely in the zillion numbers running around in banks.

It is a interesting topic you stumbled upon however I am not sure your solution is the answer

Rounding 8.55 with convert_down => 8

Since it over .5 rounding up or down should go to 9

you almost need a check just for this special case

@mbova407

Rounding 8.55 with convert_down => 8

What I mean rounding up is the default case.... eg. 4.4->4 ,4.8->5 This is rounding up that I meant

4.4->4,4.8->4 ....is rounding down that I mean..

may be my choice of words cause confusion to you...

In the old days of Numerical Control when many of the machines only understood incremental positioning and it was critical to loop to the exact same location in a program within .0001in, and the biggest integer you could possibly get was 34767, we relied on scaling and double precision to get it right. Out = SIGN(dIn) * DINT((ABS(dIn) * 10000.d0) + .5d0) / 10000.d0 (Yes that is FORTRAN)
Note that the value is scaled to the precision that you want and essentially truncated after .5 is added.

Removing the .8 from 4.8 to yield 4 is not "rounding", it is truncation.

Additionally the advice to add 0.4 is wrong and dangerous. It must not be considered a valid algorithm for dealing with rounding and truncation. All readers, please do not listen to this incorrect advice.

More rigorous logic and math need to be applied to problems of this sort than apparent but deceptively wrong "quick fixes" such as adding 0.4.

Well, satyajit's technique will work if you are only dealing with numbers with exactly one digit after the decimal point. Thus it is a very limited approach.

I see only three real world needs for converting doubles to integers:

• the classic one which you get with (int)(mynumber+0.5d)
• the bank rounding obtained with Convert.ToInt32(mynumber)
• truncating obtained by just casting to an integer

I have never encountered the need for me to decide whether 8.5 should go up or down, i.e. different from the bank rounding. Maybe satyajit should explain to us exactly what kind of rounding he is doing, i.e. the reason for why it needs to be different from the above three.

Hi..Sorry for late reply ...Let me tell you where I needed this type of approach...

I am working on a plan management tool where user has to give their estimate.. user can give estimate like 8.4,8.6 etc... It means 8 days and .4 of another day....

Now when we generate a report client needed x.4=>x , x.6=>x+1 , x.5=>x+1

So I tried conver.toint32 ...but as I told it has its problem which didnt suit my set of decimals

So we used Math.Round(8.5,MidpointRounding.AwayFromZero)

now (int)(mynumber+0.5d) is another way to do the same ....

while doing this I found this way to be very flexible...

Say tommorow my client say they need x.4=>x , x.6=>x+1 , x.5=>x which I think is not unlikely

Please let me know how will you do it if you are aware of some other better approach ?

Just writing that its just a quick fix will not help me to learn(for which I am in this forum :) )

Also @dishdy Well, satyajit's technique will work if you are only dealing with numbers with exactly one digit after the decimal point. Thus it is a very limited approach.

8.566 => 9 ... using (int)(mynumber+0.5d)

8.566 => 8 ... using (int)(mynumber+0.4d)

is it wrong ? Please let me know ...

satyajit,

So you are confirming that you are using this for numbers having only one digit after the decimal point.

Yet you are saying it is also valid for numbers like 8.566. I have never encountered a problem where all numbers in the range X.0 to X.5999999... should be rounded down to X and all numbers in the range X.6 to X.999999... should be rounded up to X+1.

@dishdy Yes ...in my project upyo 1 digit....

What I meant was it will work(8.566...) isnt it?

I am not saying that this will/will not occur in a real situation....(never mentioned too...)..

Thanks for your interest...

@eric

I want to know the correct way....Just if you write this is wrong isnt helping any 1