round(n)
Would people find it useful is there was a round(,) function. Eg. round1(x1) would round x1 to the nearest whole number and round10(x1) would round x1 to the nearest 10. It would be useful to use with the boolean functions.
round(n)
5 posts
Don't you *nearly* have thsi already
YOu have the int function right that rounds towards zero
e.g.
x = 3.1; int(x) = int(3.1) = 3
x = 3.6; int(x) = int(3.6) = 3;
Not rounding just truncating
now if you do
int(x+0.5) it is (for positive numbers) the same as your round(x)
e.g. in the above example again
x = 3.1; int(x+0.5) = int(3.1+0.5) = int(3.6) = 3
x = 3.6; int(x+0.5) = int(3.6+0.4) = int(4.1) = 4
and that is the same as round is it not.
It gets a little more complex for doing negative numbers as well in the same equation but can be done with the odd if.
edit/ and round10 is not there far off either
10*int((x/10)+0.5) or probably sightly quicker 10*int((x*0.1)+0.5)
but you can have any rounding equation.
This is all from memory and no way to check as no winamp here atm but hopefully the principles are there.
YOu have the int function right that rounds towards zero
e.g.
x = 3.1; int(x) = int(3.1) = 3
x = 3.6; int(x) = int(3.6) = 3;
Not rounding just truncating
now if you do
int(x+0.5) it is (for positive numbers) the same as your round(x)
e.g. in the above example again
x = 3.1; int(x+0.5) = int(3.1+0.5) = int(3.6) = 3
x = 3.6; int(x+0.5) = int(3.6+0.4) = int(4.1) = 4
and that is the same as round is it not.
It gets a little more complex for doing negative numbers as well in the same equation but can be done with the odd if.
edit/ and round10 is not there far off either
10*int((x/10)+0.5) or probably sightly quicker 10*int((x*0.1)+0.5)
but you can have any rounding equation.
This is all from memory and no way to check as no winamp here atm but hopefully the principles are there.
ceil(bla) rounds up, floor(bla) rounds down.
bla|0 rounds to the nearest integer
ceil(3.2356)=4
floor(6.9123)=6
3.42135|0=3
3.5555|0=4
If you want it to round to decimals you have to multiply the number by power of 10 before you round it (10 for 1 decimal, 100 for 2 decimals etc.), and then divide it by the same number after you round it
ceil(3.2356*10)/10=3.3
floor(6.9123*100)/100=6.91
((3.42135*1000)|0)/1000=3.421
bla|0 rounds to the nearest integer
ceil(3.2356)=4
floor(6.9123)=6
3.42135|0=3
3.5555|0=4
If you want it to round to decimals you have to multiply the number by power of 10 before you round it (10 for 1 decimal, 100 for 2 decimals etc.), and then divide it by the same number after you round it
ceil(3.2356*10)/10=3.3
floor(6.9123*100)/100=6.91
((3.42135*1000)|0)/1000=3.421
Sweet, thanks.
sorry didn't realise you had them functions although I didn't think of |