tags:

views:

95

answers:

7

In shell scripting how to find factorial of a number?

A: 

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

Wojtek
+4  A: 
#!/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
Vitalii Fedorenko
+3  A: 
echo 500 | dc -e '?[q]sQ[d1=Qd1-lFx*]dsFxp'
nos
+1 nice, although I have no clue why it works ;)
Felix Kling
This uses an external program with an arcane syntax to perform the computations.
Luther Blissett
@luther, external dedicated tools like bc, dc are able to calculate bigger numbers. Therefore, they are most appropriate tool for the job. There's a limit to what bash can calculate for bigger factorials.
ghostdog74
@gd: I was just informing Felix what was happening there. Understanding bash does not mean understanding the above statement.
Luther Blissett
+1  A: 

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
Luther Blissett
there's a limit for what bash can do. it breaks for bigger numbers. Maybe you have a way of producing the same results for factorial of say, for example, 500, with bash ?
ghostdog74
I'm fine with anybody using external tools to compute what they need. But the OP asked how to do this in bash, so I figured this was most likely an exercise in "bash for general programming", chapter "how to loop in bash".
Luther Blissett
+4  A: 

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
paxdiablo
Agreed. OTOH I found myself a couple of time in a forest with a hammer, a butter knife and busybox. So this should be tagged as "survival training"
Luther Blissett
+2  A: 
seq -s "*" 1 500 |bc
ghostdog74
And btw. this does not work on Mac OSX or any other system that does not come with the GNU coreutils (which seq is a part of)
Luther Blissett
then use jot on Mac OS. If any other system
ghostdog74
A: 

Here is a recursive function in Bash:

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

Of course it's quite slow and limited.

Dennis Williamson