Bubble sort in Linux/Unix shell scripting - linux

I am trying to perform bubble sort is Unix shell script. Why is my code not working?
a=(10 8 20 25 12)
for ((i=0;i<5;i++))
do
for((j=0;j<5;j++))
do
if ((${a[j]} > ${a[$((j+1))]}))
then
v=${a[$j]}
a[$j]=${a[$((j+1))]}
a[$((j+1))]=$v
fi
done
done
echo ${a[*]}
echo "end..."

I guess this is homework. therefore I don't give codes, just point out the errors in your codes:
for((j=0;j<5;j++)) then read a[j+1], here would be problem because when j=4, j+1 doesn't exist
fix that, your program will sort.

Try this:
echo "Enter size of array";
read n; #get the size of array from user.
echo "Enter the array";
read -a arr; #get the array form user eg: 2 3 4 5 6
echo "Orignal array is: ${arr[*]}"; #print orignal array
flag=1;
for (( i = 0; i < $n-1; i++ ))
do
flag=0;
for ((j = 0; j < $n-1-$i; j++ ))
do
if [[ ${arr[$j]} -gt ${arr[$j+1]} ]]
then
temp=${arr[$j]};
arr[$j]=${arr[$j+1]};
arr[$j+1]=$temp;
flag=1;
fi
done
if [[ $flag -eq 0 ]]; then
break;
fi
done
echo "Final sorted Array is ${arr[*]}";

Related

how to find sum of even no till 100 in linux

what should be condition for while before adding while loops it prints even no upto 100 but I need to print sum of even numbers
#!/bin/bash
sum=0
for((n=2;n<=100;n=n+2))
do
echo $n
while [[$n 0]] # what should be condition for while loop
do
sum= `expr sum + $n`
done
echo "sum is $sum "
done
Does this count ^_* :
kent$ seq -s + 2 2 100|bc
2550
k=0;
for i in {1..100}; do
if [[ $(( i % 2 )) == 0 ]]; then
let k=k+i
fi ; done
echo $k
Prints
2550
or this:
k=0; for i in {1..100}; do
if (( i % 2 == 0 )); then
(( k=k+i )) ; fi ; done ; echo $k
You shouldn't have the while loop at all. You're already iterating with the for loop.
#!/bin/bash
sum=0
for((n=2;n<=100;n=n+2))
do
echo $n
((sum+=n))
done
echo "sum is $sum "

Bash script to give n numbers and print even numbers

I'm trying to write a bash script for
"Given a positive integer N greater than 1, make a script to show the
even numbers between 0 and N. Ex .: The number 12 has been read. The
program must have as output: 0, 2, 4, 6, 8, 10 and 12;"
So far I did like this:
ex5.sh
#!/bin/bash
echo "Number :"
read num;
for (( i = 1; i >= $num; i++ ))
do
if [ $(($i % 2)) -eq 0 ]
then
echo $i
fi
done
When i compile, it doesn't print numbers. I couldn't find where is the problem
I tried using for-in loop as well
#!/bin/bash
echo "Number :"
read num;
for i in $num
do
if [ $(($i % 2)) -eq 0 ]
then
echo $i
fi
done
This time only print the number that i put as a input. For example, I put 4 in terminal it prints 4 as well
Many thanks
This way can make your code working:
echo -n "Number : "
read num;
for i in $(seq 0 $num)
do
if [ $(expr $i % 2) == 0 ]
then
echo $i
fi
done
bash ex5.sh
Number : 12
0
2
4
6
8
10
12
`
echo -n "Number : "
read num;
for i in $(seq 0 $num)
do
if [ $((num%2)) -eq 0 ]
then
echo $i
fi
done
`
#!/bin/bash
read -p "Number: " num
awk '{for(i=0;i<=$1;i++) if(i%2 == 0) print i}'<<<$num
./printNumbers.sh
Number: 8
0
2
4
6
8
# to get a list use printf
awk '{for(i=0;i<=$1;i++) if(i%2 == 0) printf i","}'<<<$num | sed 's/,$//'
./printNumbers.sh
Number: 8
0,2,4,6,8
#!/usr/bin/env bash
read -p 'Number: ' num
seq 0 2 "$num"
it's <= not >=
#!/bin/bash
echo "Number :"
read num;
for (( i = 1; i <= $num; i++ ))
do
if [ $(($i % 2)) -eq 0 ]
then
echo $i
fi
done

How to write script for arithmetic mean of even numbers

I have to write a script in bash where I have to provide 10 numbers to the table. Then script have to write the content out and arithmetic mean of even numbers. I did like 90% of the script, but I can't find out how to extract information about quantity of even numbers that is needed for arithmetic mean.
Here is my code:
echo "Provide data:"
i=0
for (( i = 0 ; i < 10; i++ ))
do
echo "Provide $[$i+1] number:"
read x
if [ "$x" = "" ]
then
break
else
table[$i]=$x
fi
done
echo "Provided data: ${table[*]}"
result=0
for (( i = 0 ; i < 10; i++))
do
res=$[${table[i]}%2]
if [ $res -eq 0 ]
then
echo "Number ${table[i]} is even"
result=$[$result+${table[$i]}]
fi
done
echo "SUM:$[$result]"
ignoring data input adding only odd inputs can look like :
$ cat c.sh
#!/bin/bash
declare -A xDarray
sum=0
xDarray[0 1]=1
xDarray[0 2]=3
xDarray[1 0]=2
xDarray[2 0]=4
for var in ${xDarray[#]}
do
if [ $(( $var & 1 )) == 0 ] ; then
echo $var is even
i+=1
tab[$i]=$var
sum=$(( $sum + $var))
fi
done
var=$(echo ${tab[#]} | sed 's/ / + /g' )
echo $var = $sum
result in
$ ./c.sh
2 is even
4 is even
2 + 4 = 6
$
whatever the number of data is used it would work
I let you work around your data input
Here are some suggested modifications for your script, syntax is simple.
#!/bin/bash
arr=()
for (( i=1;i<=10;i++ )); do
number=''
while [[ ! $number =~ ^[0-9]+$ ]]; do
printf "Please enter number $i:\n"
read number
done
arr+=($number)
done
printf "\nProvided numbers:"
printf " %d" "${arr[#]}"
printf "\nEven numbers:"
s=0
n=0
for x in "${arr[#]}"; do
if ! (( x % 2 )); then
printf " %d" "$x"
s=$(( s + x ))
(( n++ ))
fi
done
m=$(( s / n ))
printf "\nMean of the %d even numbers: %d / %d = %d\n" "$n" "$s" "$n" "$m"
Use an array arr to hold the input numbers, declare with arr=(), append numbers with arr+=($x), we refer only ${arr[#]} for all the items and we avoid any other complex array references, indices etc.
Every input number is tested against regular expression ^[0-9]+$ which means one or more digits (and only digits) with the =~ operator, and if this is not true, we prompt again for the same i-th input number.
Also we prefer printf for printing.
The last loop is the standard array loop, where we use again the arithmetic expansion syntax to find the even numbers, to add them to the sum and get the mean of them (result is an integer).
If you want to print a decimal result, e.g. with 2 floating points, you could use bc and printf "%f" like this:
m=$( bc <<< "scale=2; $s/$n" )
printf "%.2f" "$m"

Linux Shell-Script code is not executing as it should

Bubble Sort : (Original Code)
clear
echo "Enter size : "
read size
echo "Enter elements : "
for (( i=0; i<size; i++ ))
do
echo "Element " $i " : "
read arr[$i]
done
m=$(($n-1))
for (( i=0; i<m; i++ ))
do
cond=$((size-$i-1))
for (( j=0; j<cond; j++ ))
do
l=$(($j+1));
if [ ${a[$j]} -gt ${a[$l]} ]
then
tmp=${a[$j]}
a[$j]=${a[$l]}
a[$l]=$tmp
fi
done
done
echo "Sorted Data : "
echo ${a[*]}
This code is executing, while it takes the input and stores them in the array.
The the next thing I see is, "Sorted Array" being printed.
and no elements are printed after that.
Tried, sh -x sort.sh and the interim code isn't being executed.
I am quite new to shell script and cannot understand what's wrong.
Bubble Sort : (Corrected Code)
clear
echo "Enter size : "
read size
echo "Enter elements : "
for (( i=0; i<size; i++ ))
do
echo "Element " $i " : "
read arr[$i]
done
m=$(($size-1))
for (( i=0; i<m; i++ ))
do
cond=$(($size-$i-1))
for (( j=0; j<cond; j++ ))
do
l=$(($j+1));
if [ ${arr[$j]} -gt ${arr[$l]} ]
then
tmp=${arr[$j]}
arr[$j]=${arr[$l]}
arr[$l]=$tmp
fi
done
done
echo "Sorted Data : "
echo ${arr[*]}
Two things that are wrong in your program:
Line No. 8 - You are calling your array arr[] but later referring it as a[] . Change one of the names to match the other.
Line No. 10 - There is no $n in your program. You mean $size. Change this as well.
I think your program should work after this.
Problem 1: You've used $size and $n interchangeably.
Problem 2: You've again used the arr[] and a[] arrays interchangeably.

Bash string comparison not working

I have the following Bash function:
checkForUpdates() {
checkLatest
ret=$?
if [ $ret != 0 ]; then
return $ret
fi
count=0
for i in $(ssh $__updateuser#$__updatehost "ls $__updatepath/*${latest}*"); do
file="${i##$__updatepath}"
echo "$file" >> $__debuglog
if [ -f $__pkgpath/$file ]; then
remoteHash=$(ssh $__updateuser#$__updatehost "md5sum -b < $__updatepath/${file}")
localHash=$(md5sum -b < $__pkgpath/$file)
echo "${remoteHash:0:32} = ${localHash:0:32}" >> $__debuglog
if [ "${remoteHash:0:32}" != "${localHash:0:32}" ]; then
files[$count]=$file
count=$(($count + 1))
echo "Hashes not matched, adding $i" >> $__debuglog
fi
else
files[$count]=$file
count=$(($count + 1))
echo "$file missing" >> $__debuglog
fi
done
# Verify that the files array isn't empty.
if [ $count != 0 ]; then
return 0
else
return 33
fi
}
For some reason, the remoteHash/localHash comparison always returns true. I added the echo so that I could see the values of the hashes and they are definitely different and I can't figure out where I'm going wrong. I have tried different operators with no success and it is driving me crazy!
this isn't related to your question but more of general advice, first and most important you shouldn't parse the output of ls instead use find -print0 here's an example: http://mywiki.wooledge.org/BashFAQ/001
also consider using [[ instead of [ see: http://mywiki.wooledge.org/BashFAQ/031
now regarding your code, this part:
checkLatest
ret=$?
if [ $ret != 0 ]; then
return $ret
fi
could be written simply as:
checkLatest || return
and you don't need to keep a counter on the index of the array, if you initialize the var as an empty array like files=() you can then append elements to it with files+=("$file") you can get the count with "${#files[#]}"

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