How can I do the sum of integers with bash script I read some variables with a for and I need to do the sum.
I have written the code like this:
Read N
Sum=0
for ((i=1;i<=N;i++))
do
read number
sum=sum+number
done
echo $sum
Use the arithmetic command ((...)):
#! /bin/bash
read n
sum=0
for ((i=1; i<=n; i++)) ; do
read number
((sum+=number))
done
echo $sum
#!/bin/bash
echo "Enter number:"
read N
re='^[0-9]+$'
if ! [[ ${N} =~ ${re} ]]
then
echo "Error. It's not a number"
exit 1
fi
Sum=0
for ((i=1;i<=N;i++))
do
sum=$((${sum} + ${i}))
done
echo "${sum}"
Well, not a straight bash solution, but you can also use seq and datamash (https://www.gnu.org/software/datamash/):
#!/bin/bash
read N
seq 1 $N | datamash sum 1
It is really simple (and it has its limitations), but it works. You can use other options on seq for increments different than 1 and so on.
It is also possible to declare a variable as an integer with declare -i. Any assignment to that variable is then evaluated as an arithmetic expression:
#!/bin/bash
declare -i sum=0
read -p "Enter n: " n
for ((i=1; i<=n; i++)) ; do
read -p "Enter number #$i: " number
sum+=number #sum=sum+number would also work
done
echo "Sum: $sum"
See Bash Reference Manual for more information. Using arithmetic command ((...)) is preferred, see choroba's answer.
$ declare -i var1=1
$ var2=1
$ var1+=5
$ echo "$var1"
6
$ var2+=5
$ echo "$var2"
15
This can be a tad confusing as += behaves differently depending on the variable's attributes. It's therefore better to explicitly use ((...)) for arithmetic operations.
Related
I have this Bash script and I had a problem in line 16.
How can I take the previous result of line 15 and add
it to the variable in line 16?
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in `ls output-$i-*`; do
echo "$j"
metab=$(cat $j|grep EndBuffer|awk '{sum+=$2} END { print sum/120}') (line15)
num= $num + $metab (line16)
done
echo "$num"
done
For integers:
Use arithmetic expansion: $((EXPR))
num=$((num1 + num2))
num=$(($num1 + $num2)) # Also works
num=$((num1 + 2 + 3)) # ...
num=$[num1+num2] # Old, deprecated arithmetic expression syntax
Using the external expr utility. Note that this is only needed for really old systems.
num=`expr $num1 + $num2` # Whitespace for expr is important
For floating point:
Bash doesn't directly support this, but there are a couple of external tools you can use:
num=$(awk "BEGIN {print $num1+$num2; exit}")
num=$(python -c "print $num1+$num2")
num=$(perl -e "print $num1+$num2")
num=$(echo $num1 + $num2 | bc) # Whitespace for echo is important
You can also use scientific notation (for example, 2.5e+2).
Common pitfalls:
When setting a variable, you cannot have whitespace on either side of =, otherwise it will force the shell to interpret the first word as the name of the application to run (for example, num= or num)
num= 1 num =2
bc and expr expect each number and operator as a separate argument, so whitespace is important. They cannot process arguments like 3+ +4.
num=`expr $num1+ $num2`
Use the $(( )) arithmetic expansion.
num=$(( $num + $metab ))
See Chapter 13. Arithmetic Expansion for more information.
There are a thousand and one ways to do it. Here's one using dc (a reverse Polish desk calculator which supports unlimited precision arithmetic):
dc <<<"$num1 $num2 + p"
But if that's too bash-y for you (or portability matters) you could say
echo $num1 $num2 + p | dc
But maybe you're one of those people who thinks RPN is icky and weird; don't worry! bc is here for you:
bc <<< "$num1 + $num2"
echo $num1 + $num2 | bc
That said, there are some unrelated improvements you could be making to your script:
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in output-$i-* ; do # 'for' can glob directly, no need to ls
echo "$j"
# 'grep' can read files, no need to use 'cat'
metab=$(grep EndBuffer "$j" | awk '{sum+=$2} END { print sum/120}')
num=$(( $num + $metab ))
done
echo "$num"
done
As described in Bash FAQ 022, Bash does not natively support floating point numbers. If you need to sum floating point numbers the use of an external tool (like bc or dc) is required.
In this case the solution would be
num=$(dc <<<"$num $metab + p")
To add accumulate possibly-floating-point numbers into num.
In Bash,
num=5
x=6
(( num += x ))
echo $num # ==> 11
Note that Bash can only handle integer arithmetic, so if your AWK command returns a fraction, then you'll want to redesign: here's your code rewritten a bit to do all math in AWK.
num=0
for ((i=1; i<=2; i++)); do
for j in output-$i-*; do
echo "$j"
num=$(
awk -v n="$num" '
/EndBuffer/ {sum += $2}
END {print n + (sum/120)}
' "$j"
)
done
echo "$num"
done
I always forget the syntax so I come to Google Search, but then I never find the one I'm familiar with :P. This is the cleanest to me and more true to what I'd expect in other languages.
i=0
((i++))
echo $i;
I really like this method as well. There is less clutter:
count=$[count+1]
#!/bin/bash
read X
read Y
echo "$(($X+$Y))"
You should declare metab as integer and then use arithmetic evaluation
declare -i metab num
...
num+=metab
...
For more information, see 6.5 Shell Arithmetic.
Use the shell built-in let. It is similar to (( expr )):
A=1
B=1
let "C = $A + $B"
echo $C # C == 2
Source: Bash let builtin command
Another portable POSIX compliant way to do in Bash, which can be defined as a function in .bashrc for all the arithmetic operators of convenience.
addNumbers () {
local IFS='+'
printf "%s\n" "$(( $* ))"
}
and just call it in command-line as,
addNumbers 1 2 3 4 5 100
115
The idea is to use the Input-Field-Separator(IFS), a special variable in Bash used for word splitting after expansion and to split lines into words. The function changes the value locally to use word-splitting character as the sum operator +.
Remember the IFS is changed locally and does not take effect on the default IFS behaviour outside the function scope. An excerpt from the man bash page,
The shell treats each character of IFS as a delimiter, and splits the results of the other expansions into words on these characters. If IFS is unset, or its value is exactly , the default, then sequences of , , and at the beginning and end of the results of the previous expansions are ignored, and any sequence of IFS characters not at the beginning or end serves to delimit words.
The "$(( $* ))" represents the list of arguments passed to be split by + and later the sum value is output using the printf function. The function can be extended to add scope for other arithmetic operations also.
#!/usr/bin/bash
#integer numbers
#===============#
num1=30
num2=5
echo $(( num1 + num2 ))
echo $(( num1-num2 ))
echo $(( num1*num2 ))
echo $(( num1/num2 ))
echo $(( num1%num2 ))
read -p "Enter first number : " a
read -p "Enter second number : " b
# we can store the result
result=$(( a+b ))
echo sum of $a \& $b is $result # \ is used to espace &
#decimal numbers
#bash only support integers so we have to delegate to a tool such as bc
#==============#
num2=3.4
num1=534.3
echo $num1+$num2 | bc
echo $num1-$num2 | bc
echo $num1*$num2 |bc
echo "scale=20;$num1/$num2" | bc
echo $num1%$num2 | bc
# we can store the result
#result=$( ( echo $num1+$num2 ) | bc )
result=$( echo $num1+$num2 | bc )
echo result is $result
##Bonus##
#Calling built in methods of bc
num=27
echo "scale=2;sqrt($num)" | bc -l # bc provides support for calculating square root
echo "scale=2;$num^3" | bc -l # calculate power
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in `ls output-$i-*`; do
echo "$j"
metab=$(cat $j|grep EndBuffer|awk '{sum+=$2} END { print sum/120}') (line15)
let num=num+metab (line 16)
done
echo "$num"
done
Works on MacOS. bc is a command line calculator
#!/bin/bash
sum=0
for (( i=1; i<=5; i++ )); do
sum=$(echo "$sum + 1.1" | bc) # bc: if you want to use decimal
done
echo "Total: $sum"
Assuming there is an input:
1,2,C
We are trying to output it as
KEY=1, VAL1=2, VAL2=C
So far trying to modify from here:
Is there a way to create key-value pairs in Bash script?
for i in 1,2,C ; do KEY=${i%,*,*}; VAL1=${i#*,}; VAL2=${i#*,*,}; echo $KEY" XX "$VAL1 XX "$VAL2"; done
Output:
1 XX 2,c XX c
Not entirely sure what the pound ("#") and % here mean above, making the modification kinda hard.
Could any guru enlighten? Thanks.
I would generally prefer easier to read code, as bash can get ugly pretty fast.
Try this:
key_values.sh
#!/bin/bash
IFS=,
count=0
# $* is the expansion of all the params passed in, i.e. $1, $2, $3, ...
for i in $*; do
# '-eq' is checking for equality, i.e. is $count equal to zero.
if [ $count -eq 0 ]; then
echo -n "KEY=$i"
else
echo -n ", VAL${count}=$i"
fi
count=$(( $count + 1 ))
done
echo
Example
key_values.sh 1,2,ABC,123,DEF
Output
KEY=1, VAL1=2, VAL2=ABC, VAL3=123, VAL4=DEF
Expanding on anishsane's comment:
$ echo $1
1,2,3,4,5
$ IFS=, read -ra args <<<"$1" # read into an array
$ out="KEY=${args[0]}"
$ for ((i=1; i < ${#args[#]}; i++)); do out+=", VAL$i=${args[i]}"; done
$ echo "$out"
KEY=1, VAL1=2, VAL2=3, VAL3=4, VAL4=5
I can't seem to be able to increase the variable value by 1. I have looked at tutorialspoint's Unix / Linux Shell Programming tutorial but it only shows how to add together two variables.
I have tried the following methods but they don't work:
i=0
$i=$i+1 # doesn't work: command not found
echo "$i"
$i='expr $i+1' # doesn't work: command not found
echo "$i"
$i++ # doesn't work*, command not found
echo "$i"
How do I increment the value of a variable by 1?
You can use an arithmetic expansion like so:
i=$((i+1))
or declare i as an integer variable and use the += operator for incrementing its value.
declare -i i=0
i+=1
or use the (( construct.
((i++))
The way to use expr:
i=0
i=`expr $i + 1`
The way to use i++ (unless you're running with -e/-o errexit):
((i++)); echo $i;
Tested in gnu bash.
you can use bc as it can also do floats
var=$(echo "1+2"|bc)
These are the methods I know:
ichramm#NOTPARALLEL ~$ i=10; echo $i;
10
ichramm#NOTPARALLEL ~$ ((i+=1)); echo $i;
11
ichramm#NOTPARALLEL ~$ ((i=i+1)); echo $i;
12
ichramm#NOTPARALLEL ~$ i=`expr $i + 1`; echo $i;
13
Note the spaces in the last example, also note that's the only one that uses $i.
How would you achieve this in bash. It's a question I got asked in an interview and I could think of answers in high level languages but not in shell.
As I understand it, the real implementation of tail seeks to the end of the file and then reads backwards.
The main idea is to keep a fixed-size buffer and to remember the last lines. Here's a quick way to do a tail using the shell:
#!/bin/bash
SIZE=5
idx=0
while read line
do
arr[$idx]=$line
idx=$(( ( idx + 1 ) % SIZE ))
done < text
for ((i=0; i<SIZE; i++))
do
echo ${arr[$idx]}
idx=$(( ( idx + 1 ) % SIZE ))
done
If all not-tail commands are allowed, why not be whimsical?
#!/bin/sh
[ -r "$1" ] && exec < "$1"
tac | head | tac
Use wc -l to count the number of lines in the file. Subtract the number of lines you want from this, and add 1, to get the starting line number. Then use this with sed or awk to start printing the file from that line number, e.g.
sed -n "$start,\$p"
There's this:
#!/bin/bash
readarray file
lines=$(( ${#file[#]} - 1 ))
for (( line=$(($lines-$1)), i=${1:-$lines}; (( line < $lines && i > 0 )); line++, i-- )); do
echo -ne "${file[$line]}"
done
Based on this answer: https://stackoverflow.com/a/8020488/851273
You pass in the number of lines at the end of the file you want to see then send the file via stdin, puts the entire file into an array, and only prints the last # lines of the array.
The only way I can think of in “pure” shell is to do a while read linewise on the whole file into an array variable with indexing modulo n, where n is the number of tail lines (default 10) — i.e. a circular buffer, then iterate over the circular buffer from where you left off when the while read ends. It's not efficient or elegant, in any sense, but it'll work and avoids reading the whole file into memory. For example:
#!/bin/bash
incmod() {
let i=$1+1
n=$2
if [ $i -ge $2 ]; then
echo 0
else
echo $i
fi
}
n=10
i=0
buffer=
while read line; do
buffer[$i]=$line
i=$(incmod $i $n)
done < $1
j=$i
echo ${buffer[$i]}
i=$(incmod $i $n)
while [ $i -ne $j ]; do
echo ${buffer[$i]}
i=$(incmod $i $n)
done
This script somehow imitates tail:
#!/bin/bash
shopt -s extglob
LENGTH=10
while [[ $# -gt 0 ]]; do
case "$1" in
--)
FILES+=("${#:2}")
break
;;
-+([0-9]))
LENGTH=${1#-}
;;
-n)
if [[ $2 != +([0-9]) ]]; then
echo "Invalid argument to '-n': $1"
exit 1
fi
LENGTH=$2
shift
;;
-*)
echo "Unknown option: $1"
exit 1
;;
*)
FILES+=("$1")
;;
esac
shift
done
PRINTHEADER=false
case "${#FILES[#]}" in
0)
FILES=("/dev/stdin")
;;
1)
;;
*)
PRINTHEADER=true
;;
esac
IFS=
for I in "${!FILES[#]}"; do
F=${FILES[I]}
if [[ $PRINTHEADER == true ]]; then
[[ I -gt 0 ]] && echo
echo "==> $F <=="
fi
if [[ LENGTH -gt 0 ]]; then
LINES=()
COUNT=0
while read -r LINE; do
LINES[COUNT++ % LENGTH]=$LINE
done < "$F"
for (( I = COUNT >= LENGTH ? LENGTH : COUNT; I; --I )); do
echo "${LINES[--COUNT % LENGTH]}"
done
fi
done
Example run:
> bash script.sh -n 12 <(yes | sed 20q) <(yes | sed 5q)
==> /dev/fd/63 <==
y
y
y
y
y
y
y
y
y
y
y
y
==> /dev/fd/62 <==
y
y
y
y
y
> bash script.sh -4 <(yes | sed 200q)
y
y
y
y
Here's the answer I would give if I were actually asked this question in an interview:
What environment is this where I have bash but not tail? Early boot scripts, maybe? Can we get busybox in there so we can use the full complement of shell utilities? Or maybe we should see if we can squeeze a stripped-down Perl interpreter in, even without most of the modules that would make life a whole lot easier. You know dash is much smaller than bash and perfectly good for scripting use, right? That might also help. If none of that is an option, we should check how much space a statically linked C mini-tail would need, I bet I can fit it in the same number of disk blocks as the shell script you want.
If that doesn't convince the interviewer that it's a silly question, then I go on to observe that I don't believe in using bash extensions, because the only good reason to write anything complicated in shell script nowadays is if total portability is an overriding concern. By avoiding anything that isn't portable even in one-offs, I don't develop bad habits, and I don't get tempted to do something in shell when it would be better done in a real programming language.
Now the thing is, in truly portable shell, arrays may not be available. (I don't actually know whether the POSIX shell spec has arrays, but there certainly are legacy-Unix shells that don't have them.) So, if you have to emulate tail using only shell builtins and it's got to work everywhere, this is the best you can do, and yes, it's hideous, because you're writing in the wrong language:
#! /bin/sh
a=""
b=""
c=""
d=""
e=""
f=""
while read x; do
a="$b"
b="$c"
c="$d"
d="$e"
e="$f"
f="$x"
done
printf '%s\n' "$a"
printf '%s\n' "$b"
printf '%s\n' "$c"
printf '%s\n' "$d"
printf '%s\n' "$e"
printf '%s\n' "$f"
Adjust the number of variables to match the number of lines you want to print.
The battle-scarred will note that printf is not 100% available either. Unfortunately, if all you have is echo, you are up a creek: some versions of echo cannot print the literal string "-n", and others cannot print the literal string "\n", and even figuring out which one you have is a bit of a pain, particularly as, if you don't have printf (which is in POSIX), you probably don't have user-defined functions either.
(N.B. The code in this answer, sans rationale, was originally posted by user 'Nirk' but then deleted under downvote pressure from people whom I shall charitably assume were not aware that some shells do not have arrays.)
In my script I need to expand an interval, e.g.:
input: 1,5-7
to get something like the following:
output: 1,5,6,7
I've found other solutions here, but they involve python and I can't use it in my script.
Solution with Just Bash 4 Builtins
You can use Bash range expansions. For example, assuming you've already parsed your input you can perform a series of successive operations to transform your range into a comma-separated series. For example:
value1=1
value2='5-7'
value2=${value2/-/..}
value2=`eval echo {$value2}`
echo "input: $value1,${value2// /,}"
All the usual caveats about the dangers of eval apply, and you'd definitely be better off solving this problem in Perl, Ruby, Python, or AWK. If you can't or won't, then you should at least consider including some pipeline tools like tr or sed in your conversions to avoid the need for eval.
Try something like this:
#!/bin/bash
for f in ${1//,/ }; do
if [[ $f =~ - ]]; then
a+=( $(seq ${f%-*} 1 ${f#*-}) )
else
a+=( $f )
fi
done
a=${a[*]}
a=${a// /,}
echo $a
Edit: As #Maxim_united mentioned in the comments, appending might be preferable to re-creating the array over and over again.
This should work with multiple ranges too.
#! /bin/bash
input="1,5-7,13-18,22"
result_str=""
for num in $(tr ',' ' ' <<< "$input"); do
if [[ "$num" == *-* ]]; then
res=$(seq -s ',' $(sed -n 's#\([0-9]\+\)-\([0-9]\+\).*#\1 \2#p' <<< "$num"))
else
res="$num"
fi
result_str="$result_str,$res"
done
echo ${result_str:1}
Will produce the following output:
1,5,6,7,13,14,15,16,17,18,22
expand_commas()
{
local arg
local st en i
set -- ${1//,/ }
for arg
do
case $arg in
[0-9]*-[0-9]*)
st=${arg%-*}
en=${arg#*-}
for ((i = st; i <= en; i++))
do
echo $i
done
;;
*)
echo $arg
;;
esac
done
}
Usage:
result=$(expand_commas arg)
eg:
result=$(expand_commas 1,5-7,9-12,3)
echo $result
You'll have to turn the separated words back into commas, of course.
It's a bit fragile with bad inputs but it's entirely in bash.
Here's my stab at it:
input=1,5-7,10,17-20
IFS=, read -a chunks <<< "$input"
output=()
for chunk in "${chunks[#]}"
do
IFS=- read -a args <<< "$chunk"
if (( ${#args[#]} == 1 )) # single number
then
output+=(${args[*]})
else # range
output+=($(seq "${args[#]}"))
fi
done
joined=$(sed -e 's/ /,/g' <<< "${output[*]}")
echo $joined
Basically split on commas, then interpret each piece. Then join back together with commas at the end.
A generic bash solution using the sequence expression `{x..y}'
#!/bin/bash
function doIt() {
local inp="${#/,/ }"
declare -a args=( $(echo ${inp/-/..}) )
local item
local sep
for item in "${args[#]}"
do
case ${item} in
*..*) eval "for i in {${item}} ; do echo -n \${sep}\${i}; sep=, ; done";;
*) echo -n ${sep}${item};;
esac
sep=,
done
}
doIt "1,5-7"
Should work with any input following the sample in the question. Also with multiple occurrences of x-y
Use only bash builtins
Using ideas from both #Ansgar Wiechers and #CodeGnome:
input="1,5-7,13-18,22"
for s in ${input//,/ }
do
if [[ $f =~ - ]]
then
a+=( $(eval echo {${s//-/..}}) )
else
a+=( $s )
fi
done
oldIFS=$IFS; IFS=$','; echo "${a[*]}"; IFS=$oldIFS
Works in Bash 3
Considering all the other answers, I came up with this solution, which does not use any sub-shells (but one call to eval for brace expansion) or separate processes:
# range list is assumed to be in $1 (e.g. 1-3,5,9-13)
# convert $1 to an array of ranges ("1-3" "5" "9-13")
IFS=,
local range=($1)
unset IFS
list=() # initialize result list
local r
for r in "${range[#]}"; do
if [[ $r == *-* ]]; then
# if the range is of the form "x-y",
# * convert to a brace expression "{x..y}",
# * using eval, this gets expanded to "x" "x+1" … "y" and
# * append this to the list array
eval list+=( {${r/-/..}} )
else
# otherwise, it is a simple number and can be appended to the array
list+=($r)
fi
done
# test output
echo ${list[#]}