Ceiling of float number

ddenial · 08-08-2020, 06:46 AM

Hello All

How do I get the ceiling of the float number? I don't want float to be rounded to nearest.

For example, 11.2, 11.5 11.8 should round to its ceiling 12

I don't want traditional rounding like 11.2 = 11 and 11.8 = 12

I searched the web, but all I get is how to use a modulus trick between two numbers and finding its ceiling.

But I don't have two numbers, I only have a single float number which I derived from other complex maths. Now I want its ceiling.

The printf gives traditional rounding

Code:

$ val1=11.2
$ val2=11.8

$ printf "%.f\n" $val1
11

$ printf "%.f\n" $val2
12

For both values I want the result to be 12.

How do I do that?

Thanks

berndbausch · 08-08-2020, 06:56 AM

A bit convoluted perhaps, but should work:

Code:

echo $(($(printf "%.0f" $f)+1))

I.e. remove the fraction, then add 1.

boughtonp · 08-08-2020, 07:18 AM

You haven't given parameters as to what software you can/can't use - e.g. python and perl can both do it.

Alternatively, here's a hacky workaround that truncates zero decimals and increments any non-zero decimal part:

Code:

echo "$num" | sed -e 's/\.0*$//;s/\.[0-9]*$/ + 1/' | bc

Or if bc isn't available:

Code:

echo $(( echo $( echo "$num" | sed -e 's/\.0*$//;s/\.[0-9]*$/ + 1/' ) ))

But if you're dealing with floating point numbers, you should aim to be using a language that has native support for them.

ddenial · 08-08-2020, 08:12 AM

Quote:

Originally Posted by berndbausch

A bit convoluted perhaps, but should work:

Code:

echo $(($(printf "%.0f" $f)+1))

I.e. remove the fraction, then add 1.

Interesting. But what if I get a float with .00 as a decimal? Your solution made me think.

Code:

$ val1=11.00
$ val2=11.28
$ val3=11.68

$ echo $(($(printf "%.0f" $val1)+1))
12

$ echo $(($(printf "%.0f" $val2)+1))
12

$ echo $(($(printf "%.0f" $val3)+1))
13

It nuked for 11.00 and 11.68. Probably I should try with variable substring removal method something like this ${val3%.*} and add 1 to it.

But this also nuked for 11.00. Hmmm

ddenial · 08-08-2020, 08:14 AM

Quote:

Originally Posted by boughtonp

You haven't given parameters as to what software you can/can't use - e.g. python and perl can both do it.

Alternatively, here's a hacky workaround that truncates zero decimals and increments any non-zero decimal part:

Code:

echo "$num" | sed -e 's/\.0*$//;s/\.[0-9]*$/ + 1/' | bc

Or if bc isn't available:

Code:

echo $(( echo $( echo "$num" | sed -e 's/\.0*$//;s/\.[0-9]*$/ + 1/' ) ))

But if you're dealing with floating point numbers, you should aim to be using a language that has native support for them.

I'm writing a bash shell script and I can use bc. I don't think using bc will create a problem.

OK this one worked

Code:

$ echo "$val1" | sed -e 's/\.0*$//;s/\.[0-9]*$/ + 1/' | bc
11

$ echo "$val2" | sed -e 's/\.0*$//;s/\.[0-9]*$/ + 1/' | bc
12

$ echo "$val3" | sed -e 's/\.0*$//;s/\.[0-9]*$/ + 1/' | bc
12

dugan · 08-08-2020, 08:29 AM

There’s a bc solution here:

https://unix.stackexchange.com/a/168481

berndbausch · 08-08-2020, 09:29 AM

If we leave out the 11.00 case, perhaps the easiest method is

Code:

echo $((${val3/.*/}+1))

ondoho · 08-09-2020, 05:13 AM

Treat the number as a string, cut it at the decimal point, look at the latter value only to decide whether you need to add 1 to the former or not.

PS: do you want 11.00001 to also be rounded up to 12?