mjhjmtu

i can`t make a bash script who check if a input numbers in command line is an power of 2

input

# ./powerscript.sh xyzdf 4 8 12 -2 USAD

desired output : the desire output should be on seprated lines

4
8

because only 4 is 2^2
and 8 is 2^3

content of powerscript.sh

#!/bin/bash

function is_power_of_two () 
declare -i n=$1
(( n > 0 && (n & (n - 1)) == 0 ))


for number; do
if is_power_of_two "$number"; then
printf "%dn" "$number"
fi
done

The number is a power of 2 if its Hamming weight is exactly 1.

To calculate a number's Hamming weight is the same as calculate the number of 1s in its binary representation.

The following is a short bash script that does that:

#!/bin/bash

# loop over all numbers on the command line
# note: we don't verify that these are in fact numbers
for number do
 w=0 # Hamming weight (count of bits that are 1)
 n=$number # work on $n to save $number for later

 # test the last bit of the number, and right-shift once
 # repeat until number is zero
 while (( n > 0 )); do
 if (( (n & 1) == 1 )); then
 # last bit was 1, count it
 w=$(( w + 1 ))
 fi

 if (( w > 1 )); then
 # early bail-out: not a power of 2
 break
 fi

 # right-shift number
 n=$(( n >> 1 ))
 done

 if (( w == 1 )); then
 # this was a power of 2
 printf '%dn' "$number"
 fi
done

Testing:

$ bash script.sh xyzdf 4 8 12 -2 USAD
4
8

Note: There are more efficient ways to do this, and bash is a particularly bad choice of language for it.

There's a nice shortcut to check that a number is a power of two.

If you represent such a number in binary, it will be a single 1 followed by a string of zeroes, for instance 0b100000 for the number 32. If you subtract one from it, you'll get ones where you had the zeroes and a zero where you had the 1, for instance 0b011111 for the number 31, which is 32 - 1. If you do a bitwise and operation on these two, you'll get a zero. That property is only valid on numbers that are powers of two (and zero).

So:

function is_power_of_two () 
 declare -i n=$1
 (( n > 0 && (n & (n - 1)) == 0 ))

Use it as:

for number; do
 if is_power_of_two "$number"; then
 printf "%dn" "$number"
 fi
done

And execution output:

$ ./power2.sh 1 2 3 4 5 7 8 9 31 32 33 -2
1
2
4
8
32

Another pure bash approach

isPowerOf2 () 
 local n=$1 i=0
 for ((; n>1; n/=2, i++)); do :; done
 (($1 - (2 ** $i) == 0))

and

$ for n in 1..17; do isPowerOf2 $n && echo $n; done
1
2
4
8
16

Or looking at the octal representation of the number:

isPowerOf2() $octal =~ ^[12]0*$ ]]

Or awk perhaps

$ seq 17 | awk 'lg = log($1) / log(2) lg == int(lg)'
1
2
4
8
16

Got the factor command? Try

factor $number | sed 's/^[^:]*:|[2 ]//g;'

and test for empty result.