How do you find the factorial of a number in a Bash script?

In shell scripting how to find fac开发者_如何转开发torial of a number?

seq -s "*" 1 500 |bc

You don't do it in bash. Intelligent people don't try to cut down trees with a fish, so my advice is to try and use the right tool for the job.

You can use, for example, bc to do it thus:

pax> echo 'define f(x) {if (x>1){return x*f(x-1)};return 1}
           f(6)' | bc
720
pax> echo 'define f(x) {if (x>1){return x*f(x-1)};return 1}
           f(500)' | BC_LINE_LENGTH=99999 bc
12201368259911100687012387854230469262535743428031928421924135883858
45373153881997605496447502203281863013616477148203584163378722078177
20048078520515932928547790757193933060377296085908627042917454788242
49127263443056701732707694610628023104526442188787894657547771498634
94367781037644274033827365397471386477878495438489595537537990423241
06127132698432774571554630997720278101456108118837370953101635632443
29870295638966289116589747695720879269288712817800702651745077684107
19624390394322536422605234945850129918571501248706961568141625359056
69342381300885624924689156412677565448188650659384795177536089400574
52389403357984763639449053130623237490664450488246650759467358620746
37925184200459369692981022263971952597190945217823331756934581508552
33282076282002340262690789834245171200620771464097945611612762914595
12372299133401695523638509428855920187274337951730145863575708283557
80158735432768888680120399882384702151467605445407663535984174430480
12893831389688163948746965881750450692636533817505547812864000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000

#!/bin/bash
counter=$1 #first argument
factorial=1
while [ $counter -gt 0 ] #while counter > 0
do
   factorial=$(( $factorial * $counter ))
   counter=$(( $counter - 1 ))
done
echo $factorial

echo 500 | dc -e '?[q]sQ[d1=Qd1-lFx*]dsFxp'

10! in bash:

f=1; for k in {1..10}; do f=$[$k * $f] ; done; echo $f

or here in a step by step fashion:

$ t=$(echo {1..10})
$ echo $t
1 2 3 4 5 6 7 8 9 10
$ t=${t// /*}
$ echo $t
1*2*3*4*5*6*7*8*9*10
$ echo $[$t]
3628800

echo Enter Number

read num
fact=1
for ((i=1;i<=num;i++))
do
fact=$(($fact*$i))
done
echo $fact

Here is a recursive function in Bash:

factorial () { 
    if (($1 == 1))
    then
        echo 1
        return
    else
        echo $(( $( factorial $(($1 - 1)) ) * $1 ))
    fi
}

Of course it's quite slow and limited.

There are a number of instructive examples on Rosetta Code.

Here's one I found particularly useful:

function factorial {
  typeset n=$1
  (( n < 2 )) && echo 1 && return
  echo $(( n * $(factorial $((n-1))) ))
}

seq -s* `dd`|bc

I believe this is the shortest way of accomplishing the task.

Please use this script to find factorial in Bash,

#!/bin/bash

num=$1
fact=1
for((i=1; i<=$num; i++))
do
        let fact=fact*i
done

echo "Factorial is $fact"

You have to use loop, see this link: http://ubuntuforums.org/showthread.php?t=1349272

I almost completely agree with Vitalii Fedorenko, I would just like paxdiablo suggests use bc, here's the code from Vitalii Fedorenko but modified to use bc.

#!/bin/bash
counter=$1
output=1
while [ $counter -gt 1 ] #while counter > 1 (x*1=x)
do
        output=$(echo "$output * $counter" | bc)
        counter=$(($counter - 1))
done
#remove newlines and '\' from output
output=$(echo "$output" | tr -d '\' | tr -d '\n')
echo "$output"
exit

This method is better because bc allows you to use strings, instead of integers, making it possible for you to calculate much bigger numbers.

I apologize if I haven't used tr correctly, I am not very familiar with it.

You can use:

seq -s "*" 1 10 | sed 's/*$//g' |bc

on a mac