I need to perform a calculation(addition/multiplication) using the command line input.
For an example: I'm executing the below ./calculation.sh 1 2 3 4 5. It has to sum up the output as 15. Any idea to this ? I've tried with the below logic but couldn't make it.
set -x
while [ $# -gt 0 ]
do
expr $1 + 1
shift
done
OUTPUT=0
for i in $*; do
OUTPUT=$(($OUTPUT + $i))
done
echo $OUTPUT
You need to make use of a variable to save the result of expr. Moreover, +1 doesn't seem to make much sense. You probably wanted to replace that with the variable itself.
You need to print the variable at the end.
Try:
set -x
res=0
while [ $# -gt 0 ]
do
res=`expr $1 + $res`
shift
done
echo $res
Try
set -x
sum=0
while [ $# -gt 0 ]
do
sum=$(expr "$sum" + "$1")
shift
done
echo "sum: $sum"
And it's simpler in bash:
sum=0
for i; do
(( sum += i ))
done
echo "sum: $sum"
Related
n=20
x=3
count=0
flag=0
i=1
declare -a arr[n+1]
for (( j=0;j<=n;j++ ))
do
arr+=(0)
done
#echo "${arr[#]}"
while [[ $count -ne $n ]]
do
if [[ $i -le $n ]]
then
if [[ ${arr[$i]} -eq '0' ]]
then
echo "Value is ${arr[$i]}"
#${arr[$(i-1)]}= (( ${arr[$i-1]++} ))
${arr[$i]}+=${arr[$i]}
echo " "
#echo -n "${arr[$i]}"
echo -n " $i"
count=$(( count+1 ))
i=$(( i+1+x ))
else
i=$(( i+1 ))
fi
else
i=$(( i-n ))
flag=$(( flag+1 ))
fi
done
echo " "
echo "No of round : $flag"
This is the whole code, I've tried to print numbers that follows this: n=20 is the number of elements and x=3 is the number that we have to avoid. For example,
20
3
1,5,9,13,17,2,6,10,14,18,3,7,11,15,19,4,8,12,16,20,
3
But, the problem is that my second if condition is not fulfilling, if ignores the condition. Above example is for the C++, but in bash script, 2nd if statement isn't working. This can be because syntax is wrong. So can you please help me to find the mistakes.
Output of the above code:
output
${arr[$i]}+=${arr[$i]}
This is incorrect. $ should not be used when you assign the value.
If you want to double the value, replace this string with the following:
arr[$i]=$(( ${arr[$i]} + ${arr[$i]} ))
Or what you want to do there?
I have been trying to resolve an issue where my loop's count should decrease, however nothing is working. I need to create a while loop that will read over a given amount of times. For instance, if I enter in "files.txt -a 3" in the terminal, I need my loop to repeat "Enter in a string: " 3 times. With my code below, I am only able to get it to loop once. I am not to sure where to put the counter and I can say that I have put it everywhere. Inside the if statement, in inside of the for loop, and inside the while loop but none seem to work. The number that the user will put is held in the $count variable.
#!/bin/bash
if ["$1" = "-a" ]
then
read in user String and save into file
fi
while [ "$count" > 0 ]
do
for i in $count
do
if [ "-a" ]
then
read -p "Enter in a string: " userSTR
echo userSTR >> files.txt
count=$(($count - 1))
fi
done
done
For conditional expression you need to use [[ expression ]], e.g. this will loop four times:
count=4
while [[ $count > 0 ]] ; do
echo "$count"
count=$(( $count - 1 ))
done
To fetch the count from the command-line argument, you could replace the assignment count=4 above with the following, parsing the command-line arguments:
if [ $# -lt 2 ] ; then
echo "Usage: $0 -a [count]"
exit 1
fi
if [ "$1" = "-a" ] ; then
shift
count=$1
fi
How to write a script which will receive a list of parameters and output the number that is the largest. If no parameters are supplied, output an error message.
I wrote the following code to check if no parameters are supplied, output an error message.
#!/bin/bash
if [ "$#" -eq "0" ]
then
echo "No arugments supplied"
else
echo "$# Parameter"
But I dont know how to continue...
Keep a current max. Loop over the input, updating max if necessary. At the end you'll have the global maximum.
Untested:
#!/usr/bin/bash
if [ $# -eq 0 ]; then
echo "Usage: $0 NUMBERS" >&2
exit 1;
fi
max="$1"
shift
while [ $# -gt 0 ]; do
if [ "$1" -gt "$max" ]; then
max="$1"
fi
shift
done
echo "$max"
Use sort -n (numeric) and -r (reverse) and then just pick the first line of the output -- like
#!/bin/bash
if [ "$#" -eq "0" ]
then
echo "No arugments supplied"
else
echo "$# Parameter"
for i in $*; do echo ${i}; done | sort -nr | head -1
fi
Now the only problem you are facing is when the the input (the arguments) are not numbers -- but you didn't say anything about what should happen then.
Here's a pseudocode you could implement:
save the first param in a variable called max
loop over the params
if the param is greater than max, update max
print max
Here's an example loop that prints all parameters:
for num; do
echo $num
done
And here's an example of comparing values:
if (( num > max )); then
echo $num is greater than $max
fi
This should be more than enough help to complete your homework.
Well since others have already gave you the actual solution, here's mine too:
#!/bin/bash
if (( $# == 0 )); then
echo "No arugments supplied"
exit 1
fi
max=$1
for num; do
if (( num > max )); then
max=$num
fi
done
echo $max
I am trying to write a while loop to determine the number is being given to count down to 0. Also, if there's no argument given, must display "no parameters given.
Now I have it counting down but the last number is not being 0 and as it is counting down it starts with the number 1. I mush use a while loop.
My NEW SCRIPT.
if [ $# -eq "0" ] ;then
echo "No paramters given"
else
echo $#
fi
COUNT=$1
while [ $COUNT -gt 0 ] ;do
echo $COUNT
let COUNT=COUNT-1
done
echo Finished!
This is what outputs for me.
sh countdown.sh 5
1
5
4
3
2
1
Finished!
I need it to reach to 0
#Slizzered has already spotted your problem in a comment:
You need operator -ge (greater than or equal) rather than -gt (greater than) in order to count down to 0.
As for why 1 is printed first: that's simply due to the echo $# statement before the while loop.
If you're using bash, you could also consider simplifying your code with this idiomatic reformulation:
#!/usr/bin/env bash
# Count is passed as the 1st argument.
# Abort with error message, if not given.
count=${1?No parameters given}
# Count down to 0 using a C-style arithmetic expression inside `((...))`.
# Note: Increment the count first so as to simplify the `while` loop.
(( ++count ))
while (( --count >= 0 )); do
echo $count
done
echo 'Finished!'
${1?No parameters given} is an instance of shell parameter expansion
bash shell arithmetic is documented here.
You should also validate the variable before using it in an arithmetic context. Otherwise, a user can construct an argument that will cause the script to run in an infinite loop or hit the recursion limit and segfault.
Also, don't use uppercase variable names since you risk overriding special shell variables and environment variables. And don't use [ in bash; prefer the superior [[ and (( constructs.
#!/usr/bin/env bash
shopt -s extglob # enables extended globs
if (( $# != 1 )); then
printf >&2 'Missing argument\n'
exit 1
elif [[ $1 != +([0-9]) ]]; then
printf >&2 'Not an acceptable number\n'
exit 2
fi
for (( i = $1; i >= 0; i-- )); do
printf '%d\n' "$i"
done
# or if you insist on using while
#i=$1
#while (( i >= 0 )); do
# printf '%d\n' "$((i--))"
#done
Your code is far from being able to run. So, I don't know where to start to explain. Let's take this small script:
#!/bin/sh
die() {
echo $1 >&2
exit 1;
}
test -z "$1" && die "no parameters given"
for i in $(seq $1 -1 0); do
echo "$i"
done
The main part is the routine seq which does what you need: counting from start value to end value (with increment in between). The start value is $1, the parameter to our script, the increment is -1.
The test line tests whether there is a parameter on the command line - if not, the script ends via the subroutine die.
Hth.
There are a number of ways to do this, but the general approach is to loop from the number given to an ending number decrementing the loop count with each iteration. A C-style for loop works as well as anything. You will adjust the sleep value to get the timing you like. You should also validate the required number and type of input your script takes. One such approach would be:
#!/bin/bash
[ -n "$1" ] || {
printf " error: insufficient input. usage: %s number (for countdown)\n" "${0//*\//}"
exit 1
}
[ "$1" -eq "$1" >/dev/null 2>&1 ] || {
printf " error: invalid input. number '%s' is not an integer\n" "$1"
exit 1
}
declare -i cnt=$(($1))
printf "\nLaunch will occur in:\n\n"
for ((i = cnt; i > 0; i--)); do
printf " %2s\n" "$i"
sleep .5
done
printf "\nFinished -- blastoff!\n\n"
exit 0
Output
$ bash ./scr/tmp/stack/countdown.sh 10
Launch will occur in:
10
9
8
7
6
5
4
3
2
1
Finished -- blastoff!
Your Approach
Your approach is fine, but you need to use the value of COUNT $COUNT in your expression. You also should declare -i COUNT=$1 to tell the shell to treat it as an integer:
#!/bin/bash
if [ $# -eq "0" ] ;then
echo "No paramters given"
else
echo -e "\nNumber of arguments: $#\n\n"
fi
declare -i COUNT=$1
while [ $COUNT -gt 0 ] ;do
echo $COUNT
let COUNT=$COUNT-1
done
echo -e "\nFinished!\n"
Sorry about bits and snippit of information
So I am writing an average shell script program
so if use inputs
echo 1 3, .... | sh get_number
I would have to pull the numbers seperated by spaces from echo to be
var1 = 1, var2= 3, etc.
I tried
#!/bin/sh
sum=0
for i in $*
do
sum=`expr $sum + $i`
done
avg=`expr $sum / $n`
echo Average=$avg
but doesnt work....
do I include a read here?
also how would I do
sh get_number <file1>, <file2>... to grab numbers in them and sum them
in shell script?
Thanks
Sounds like you are looking for the read shell builtin:
% echo "1 2 3 4" | read a b stuff
% echo $b
2
% echo $stuff
3 4
To fix up your code:
for i in $*; do
sum=$(( sum + i ))
n=$(( n + 1 ))
done
echo "Average=$(( sum / n ))"
#!/bin/sh
while [ $# -gt 0 ]; do
(( i++ ))
(( sum += $1 ))
shift
done
echo "Average=$(( sum/i ))"
Note: This fails in dash which is the closest shell I could find to a real sh.
An example of reading values from files passed as command line arguments or from lines read from stdin:
add_to_sum() {
set $*
while [ $# -gt 0 ]; do
I=`expr $I + 1`
SUM=`expr $SUM + $1`
shift
done
}
I=0
SUM=0
if [ $# -gt 0 ]; then
# process any arguments on the command line
while [ $# -gt 0 ]; do
FILE=$1
shift
while read LINE; do
add_to_sum "$LINE"
done < "$FILE"
done
else
# if no arguments on the command line, read from stdin
while read LINE; do
add_to_sum "$LINE"
done
fi
# be sure not to divide by zero
[ $I -gt 0 ] && echo Average=`expr $SUM / $I`