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A quick example of what I want using bash scripting:

#!/bin/bash
echo "Insert the price you want to calculate:"
read float
echo "This is the price without taxes:"
echo "scale=2; $float/1.18" |bc -l
read -p "Press any key to continue..."
bash scriptname.sh

Assuming that the price is: 48.86 The answer will be:41.406779661 (41.40 actually because I'm using scale=2;)

My Question is: How I round the second decimal to show the answer in this way?: 41.41

share|improve this question
    
I find it weird because "printf "%0.2f\n" 41.445" does now work but "printf "%0.2f\n" 41.435 and printf "%0.2f\n" 41.455" do. Even your own case works (On 12.04) but not with the .445 –  Luis Alvarado Aug 24 '12 at 16:44
1  
IMHO, nobody has answered this question satisfactorily, perhaps because bc cannot achieve what is being requested (or at least the question I was asking when I found this post), which is how to round decimals using bc (that happens to be called by bash). –  Adam Katz May 21 at 22:51

5 Answers 5

up vote 12 down vote accepted

A bash round function:

round()
{
echo $(printf %.$2f $(echo "scale=$2;(((10^$2)*$1)+0.5)/(10^$2)" | bc))
};

Used in your code example:

#!/bin/bash
# the function "round()" was taken from 
# http://stempell.com/2009/08/rechnen-in-bash/

# the round function:
round()
{
echo $(printf %.$2f $(echo "scale=$2;(((10^$2)*$1)+0.5)/(10^$2)" | bc))
};

echo "Insert the price you want to calculate:"
read float
echo "This is the price without taxes:"
#echo "scale=2; $float/1.18" |bc -l
echo $(round $float/1.18 2);
read -p "Press any key to continue..."

Good luck :o)

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1  
Right now, this is the nicest way to do it, thanks for your knowledge. –  blackedx Aug 24 '12 at 16:51
    
btw, if the number is negative, we have to use -0.5 –  Aquarius Power Jun 4 '14 at 4:47
    
Couldn't get the example to work on the testing console! "Debugging" a little revealed that for e. g. $2 = 3 I had to use echo $(env printf %.3f $(echo "scale=3;((1000*$1)+0.5)/1000" | bc)). Mind the env before printf! This'll teach ya again that it's always important to understand what you're copy-pasting from elsewhere. –  syntaxerror Oct 23 '14 at 23:04
    
@Aquarius Power thanks for the inspiration! I've now forked a version of the above script which will work with both negative and positive numbers. –  syntaxerror Oct 23 '14 at 23:48
    
This is unnecessarily complicated and does all kinds of extra arithmetic to arrive at the simpler answers provided by migas and zuberuber. –  Adam Katz May 21 at 22:46

Bash/awk rounding:

echo "23.49" | awk '{printf("%d\n",$1 + 0.5)}'  

If you have python you can use something like this:

echo "4.678923" | python -c "print round(float(raw_input()))"
share|improve this answer
    
Thanks for the tip but it doesn't solve what I need. I have to do it just by using bash... –  blackedx Aug 24 '12 at 14:39
    
The python command is more readable and good for quick scripts. Also supports arbitrary digit rounding by adding e.g. ", 3" to the round function. Thanks –  Jordan Trudgett Nov 13 '13 at 2:07
    
echo "$float" |awk '{printf "%.2f", $1/1.18}' will perform the question's requested math to the requested percision of hundredths. That's as much "using bash" as the bc call in the question. –  Adam Katz May 21 at 22:41

Simplest solution:

printf %.2f $(echo "$float/1.18" | bc -l)
share|improve this answer

I'm still looking for a pure bc answer to how to round just one value within a function, but here's a pure bash answer:

#!/bin/bash

echo "Insert the price you want to calculate:"
read float
echo "This is the price without taxes:"

embiggen() {
  local int precision fraction=""
  if [ "$1" != "${1#*.}" ]; then  # there is a decimal point
    fraction="${1#*.}"       # just the digits after the dot
  fi
  int="${1%.*}"              # the float as a truncated integer
  precision="${#fraction}"   # the number of fractional digits
  echo $(( 10**10 * $int$fraction / 10**$precision ))
}

# round down if negative
if [ "$float" != "${float#-}" ]
  then round="-5000000000"
  else round="5000000000"
fi

# calculate rounded answer (sans decimal point)
answer=$(( ( `embiggen $float` * 100 + $round ) / `embiggen 1.18` ))

int=${answer%??}  # the answer as a truncated integer

echo $int.${answer#$int}  # reassemble with correct precision

read -p "Press any key to continue..."

Basically, this carefully extracts the decimals, multiplies everything by 100 billion (10¹⁰, 10**10 in bash), adjusts for precision and rounding, performs the actual division, divides back to the appropriate magnitude, and then reinserts the decimal.

Step-by-step:

The embiggen() function assigns the truncated integer form of its argument to $int and saves the numbers after the dot in $fraction. The number of fractional digits is noted in $precision. The math multiplies 10¹⁰ by the concatenation of $int and $fraction and then adjusts that to match the precision (e.g. embiggen 48.86 becomes 10¹⁰ × 4886 / 100 and returns 488600000000 which is 488,600,000,000).

We want a final precision of hundredths, so we multiply the first number by 100, add 5 for rounding purposes, and then divide the second number. This assignment of $answer leaves us at a hundred times the final answer.

Now we need to add the decimal point. We assign a new $int value to $answer excluding its final two digits, then echo it with a dot and the $answer excluding the $int value that's already taken care of. (Never mind the syntax highlighting bug that makes this appear like a comment)

(Bashism: exponentiation is not POSIX, so this is a bashism. A pure POSIX solution would require loops to add zeros rather than using powers of ten. Also, "embiggen" is a perfectly cromulant word.)


One of the main reasons I use zsh as my shell is that it supports floating point math. The solution to this question is quite straightforward in zsh:

printf %.2f $((float/1.18))

(I'd love to see somebody add a comment to this answer with the trick to enabling floating point arithmetic in bash, but I'm pretty sure that such a feature doesn't yet exist.)

share|improve this answer
#!/bin/bash
# - loosely based on the function "round()", taken from 
# http://stempell.com/2009/08/rechnen-in-bash/

# - inspired by user85321 @ askubuntu.com (original author)
#   and Aquarius Power

# the round function (alternate approach):

round2()
{
    v=$1
    vorig=$v
    # if negative, negate value ...
    (( $(bc <<<"$v < 0") == 1 )) && v=$(bc <<<"$v * -1")
    r=$(bc <<<"scale=$3;(((10^$3)*$v/$2)+0.5)/(10^$3)")

    # ... however, since value was only negated to get correct rounding, we 
    # have to add the minus sign again for the resulting value ...

    (( $(bc <<< "$vorig < 0") == 1 )) && r=$(bc <<< "$r * -1")
    env printf %.$3f $r
};

echo "Insert the price you want to calculate:"
read float
echo "This is the price without taxes:"
round2 $float 1.18 2
echo && read -p "Press any key to continue..."

It is actually simple: there is no need to explicitly add a hardcoded "-0.5" variant for negative numbers. Mathematically spoken, we'll just compute the absolute value of the argument and still add 0.5 as we normally would. But since we (unfortunately) have no built-in abs() function at our disposal (unless we code one), we will simply negate the argument if it's negative.

Besides, it proved very cumbersome to work with the quotient as a parameter (since for my solution, I must be able to access the dividend and divisor separately). This is why my script has an additional third parameter.

share|improve this answer
    
@muru about your recent edits: herestrings instead of echo .. |, shell arithmetic, what is the point of echo $()? This is easy to explain: your here strings will always require a writable /tmp directory! And my solution will also work on a read-only environment, e. g. an emergency root shell where / is not always writable by default. So there was a good reason why I coded it that way. –  syntaxerror Apr 17 at 14:48
    
Herestrings, I agree. But when is echo $() ever needed? That and indentation prompted my edits, the herestrings just happened. –  muru Apr 17 at 14:54
    
@muru Well the most pointless case of echo $() that was overdue to get fixed was in fact the penultimate line, lol. Thanks for the heads-up. At least this one does look much better now, I can't deny. Anyways, I loved the logic in this. Lots of boolean stuff, less superfluous "if"s (which surely would blow up the code by 50 percent). –  syntaxerror Apr 17 at 14:58

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