Linux Shell Scripting recursive exponentation











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I am writing a script that takes 2 numbers as an input and uses recursion to power on number to the power of the other, simple exponentiation. However I am new to scripting and cannot figure out where my syntax is errored here.



Here is the script



#!/bin/bash
echo "Enter number: "
read number
echo "Enter power: "
read power

echo "Powering $number to power of $power!"

exp () {

if [ $2 = 1 ]
then
return $1
fi

return $1 * $(exp $1 $2-1 )

}

result=$(exp $number, $power)

echo "Result: $result"


Currently, it kind of freezes, im not sure if I use the parameters correctly (in terms of syntax).










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  • This might help: How to debug a bash script?
    – Cyrus
    Nov 11 at 11:01










  • This might help: return can only return values from 0 to 255.
    – Cyrus
    Nov 11 at 11:03










  • @Cyrus That's what I got in bash. In dash, I was able to get 2^30 (but no higher powers of two. Bad idea to use return values in any case.
    – PSkocik
    Nov 11 at 11:23















up vote
2
down vote

favorite












I am writing a script that takes 2 numbers as an input and uses recursion to power on number to the power of the other, simple exponentiation. However I am new to scripting and cannot figure out where my syntax is errored here.



Here is the script



#!/bin/bash
echo "Enter number: "
read number
echo "Enter power: "
read power

echo "Powering $number to power of $power!"

exp () {

if [ $2 = 1 ]
then
return $1
fi

return $1 * $(exp $1 $2-1 )

}

result=$(exp $number, $power)

echo "Result: $result"


Currently, it kind of freezes, im not sure if I use the parameters correctly (in terms of syntax).










share|improve this question






















  • This might help: How to debug a bash script?
    – Cyrus
    Nov 11 at 11:01










  • This might help: return can only return values from 0 to 255.
    – Cyrus
    Nov 11 at 11:03










  • @Cyrus That's what I got in bash. In dash, I was able to get 2^30 (but no higher powers of two. Bad idea to use return values in any case.
    – PSkocik
    Nov 11 at 11:23













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am writing a script that takes 2 numbers as an input and uses recursion to power on number to the power of the other, simple exponentiation. However I am new to scripting and cannot figure out where my syntax is errored here.



Here is the script



#!/bin/bash
echo "Enter number: "
read number
echo "Enter power: "
read power

echo "Powering $number to power of $power!"

exp () {

if [ $2 = 1 ]
then
return $1
fi

return $1 * $(exp $1 $2-1 )

}

result=$(exp $number, $power)

echo "Result: $result"


Currently, it kind of freezes, im not sure if I use the parameters correctly (in terms of syntax).










share|improve this question













I am writing a script that takes 2 numbers as an input and uses recursion to power on number to the power of the other, simple exponentiation. However I am new to scripting and cannot figure out where my syntax is errored here.



Here is the script



#!/bin/bash
echo "Enter number: "
read number
echo "Enter power: "
read power

echo "Powering $number to power of $power!"

exp () {

if [ $2 = 1 ]
then
return $1
fi

return $1 * $(exp $1 $2-1 )

}

result=$(exp $number, $power)

echo "Result: $result"


Currently, it kind of freezes, im not sure if I use the parameters correctly (in terms of syntax).







linux bash shell recursion command-line






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 11 at 10:55









student126

135




135












  • This might help: How to debug a bash script?
    – Cyrus
    Nov 11 at 11:01










  • This might help: return can only return values from 0 to 255.
    – Cyrus
    Nov 11 at 11:03










  • @Cyrus That's what I got in bash. In dash, I was able to get 2^30 (but no higher powers of two. Bad idea to use return values in any case.
    – PSkocik
    Nov 11 at 11:23


















  • This might help: How to debug a bash script?
    – Cyrus
    Nov 11 at 11:01










  • This might help: return can only return values from 0 to 255.
    – Cyrus
    Nov 11 at 11:03










  • @Cyrus That's what I got in bash. In dash, I was able to get 2^30 (but no higher powers of two. Bad idea to use return values in any case.
    – PSkocik
    Nov 11 at 11:23
















This might help: How to debug a bash script?
– Cyrus
Nov 11 at 11:01




This might help: How to debug a bash script?
– Cyrus
Nov 11 at 11:01












This might help: return can only return values from 0 to 255.
– Cyrus
Nov 11 at 11:03




This might help: return can only return values from 0 to 255.
– Cyrus
Nov 11 at 11:03












@Cyrus That's what I got in bash. In dash, I was able to get 2^30 (but no higher powers of two. Bad idea to use return values in any case.
– PSkocik
Nov 11 at 11:23




@Cyrus That's what I got in bash. In dash, I was able to get 2^30 (but no higher powers of two. Bad idea to use return values in any case.
– PSkocik
Nov 11 at 11:23












1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










You need $(( )) to force arithmetic evaluation. Then you can do it with return values:



number=2 power=7
exp () {
if [ $2 -eq 1 ]; then return $1; fi
exp $1 $(($2-1))
return $(($1 * $?))
}
exp $number $power; result=$?
echo "Result: $result"


but it's a bad idea, because shells kind of reserve nonzero return values to communicate failure (e.g. the above solution
will "break" set -e).



More idiomatically, you can use stdout:



set -e
number=2 power=7
exp () {
if [ $2 -eq 1 ]; then echo $1; return; fi
echo $(($1 * $(exp $1 $(($2-1)) ) ))
}
result=$(exp $number $power)
echo "Result: $result"


but that's kind of inefficient with all the subshells.
Best to avoid the recursion and simply loop:



number=2 power=7
exp () {
local res=1 i=0;
while [ $i -lt $2 ]; do res=$((res*$1)); i=$((i+1)); done
echo $res
}
exp $number $power





share|improve this answer























  • @student126: Please take a look at: [What should I do when someone answers my question?](stackoverflow.com/help/someone-answers
    – Cyrus
    Nov 11 at 11:48











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










You need $(( )) to force arithmetic evaluation. Then you can do it with return values:



number=2 power=7
exp () {
if [ $2 -eq 1 ]; then return $1; fi
exp $1 $(($2-1))
return $(($1 * $?))
}
exp $number $power; result=$?
echo "Result: $result"


but it's a bad idea, because shells kind of reserve nonzero return values to communicate failure (e.g. the above solution
will "break" set -e).



More idiomatically, you can use stdout:



set -e
number=2 power=7
exp () {
if [ $2 -eq 1 ]; then echo $1; return; fi
echo $(($1 * $(exp $1 $(($2-1)) ) ))
}
result=$(exp $number $power)
echo "Result: $result"


but that's kind of inefficient with all the subshells.
Best to avoid the recursion and simply loop:



number=2 power=7
exp () {
local res=1 i=0;
while [ $i -lt $2 ]; do res=$((res*$1)); i=$((i+1)); done
echo $res
}
exp $number $power





share|improve this answer























  • @student126: Please take a look at: [What should I do when someone answers my question?](stackoverflow.com/help/someone-answers
    – Cyrus
    Nov 11 at 11:48















up vote
1
down vote



accepted










You need $(( )) to force arithmetic evaluation. Then you can do it with return values:



number=2 power=7
exp () {
if [ $2 -eq 1 ]; then return $1; fi
exp $1 $(($2-1))
return $(($1 * $?))
}
exp $number $power; result=$?
echo "Result: $result"


but it's a bad idea, because shells kind of reserve nonzero return values to communicate failure (e.g. the above solution
will "break" set -e).



More idiomatically, you can use stdout:



set -e
number=2 power=7
exp () {
if [ $2 -eq 1 ]; then echo $1; return; fi
echo $(($1 * $(exp $1 $(($2-1)) ) ))
}
result=$(exp $number $power)
echo "Result: $result"


but that's kind of inefficient with all the subshells.
Best to avoid the recursion and simply loop:



number=2 power=7
exp () {
local res=1 i=0;
while [ $i -lt $2 ]; do res=$((res*$1)); i=$((i+1)); done
echo $res
}
exp $number $power





share|improve this answer























  • @student126: Please take a look at: [What should I do when someone answers my question?](stackoverflow.com/help/someone-answers
    – Cyrus
    Nov 11 at 11:48













up vote
1
down vote



accepted







up vote
1
down vote



accepted






You need $(( )) to force arithmetic evaluation. Then you can do it with return values:



number=2 power=7
exp () {
if [ $2 -eq 1 ]; then return $1; fi
exp $1 $(($2-1))
return $(($1 * $?))
}
exp $number $power; result=$?
echo "Result: $result"


but it's a bad idea, because shells kind of reserve nonzero return values to communicate failure (e.g. the above solution
will "break" set -e).



More idiomatically, you can use stdout:



set -e
number=2 power=7
exp () {
if [ $2 -eq 1 ]; then echo $1; return; fi
echo $(($1 * $(exp $1 $(($2-1)) ) ))
}
result=$(exp $number $power)
echo "Result: $result"


but that's kind of inefficient with all the subshells.
Best to avoid the recursion and simply loop:



number=2 power=7
exp () {
local res=1 i=0;
while [ $i -lt $2 ]; do res=$((res*$1)); i=$((i+1)); done
echo $res
}
exp $number $power





share|improve this answer














You need $(( )) to force arithmetic evaluation. Then you can do it with return values:



number=2 power=7
exp () {
if [ $2 -eq 1 ]; then return $1; fi
exp $1 $(($2-1))
return $(($1 * $?))
}
exp $number $power; result=$?
echo "Result: $result"


but it's a bad idea, because shells kind of reserve nonzero return values to communicate failure (e.g. the above solution
will "break" set -e).



More idiomatically, you can use stdout:



set -e
number=2 power=7
exp () {
if [ $2 -eq 1 ]; then echo $1; return; fi
echo $(($1 * $(exp $1 $(($2-1)) ) ))
}
result=$(exp $number $power)
echo "Result: $result"


but that's kind of inefficient with all the subshells.
Best to avoid the recursion and simply loop:



number=2 power=7
exp () {
local res=1 i=0;
while [ $i -lt $2 ]; do res=$((res*$1)); i=$((i+1)); done
echo $res
}
exp $number $power






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 12 at 19:06

























answered Nov 11 at 11:20









PSkocik

31.7k54569




31.7k54569












  • @student126: Please take a look at: [What should I do when someone answers my question?](stackoverflow.com/help/someone-answers
    – Cyrus
    Nov 11 at 11:48


















  • @student126: Please take a look at: [What should I do when someone answers my question?](stackoverflow.com/help/someone-answers
    – Cyrus
    Nov 11 at 11:48
















@student126: Please take a look at: [What should I do when someone answers my question?](stackoverflow.com/help/someone-answers
– Cyrus
Nov 11 at 11:48




@student126: Please take a look at: [What should I do when someone answers my question?](stackoverflow.com/help/someone-answers
– Cyrus
Nov 11 at 11:48


















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