How do I iterate over a range of numbers in Bash when the range is given by a variable?
I know I can do this (called "sequence expression" in the Bash documentation):
for i in {1..5}; do echo $i; done
Which gives:
1
2
3
4
5
Yet, how can I replace either of the range endpoints with a variable? This doesn't work:
END=5
for i in {1..$END}; do echo $i; done
Which prints:
{1..5}
for i in $(seq 1 $END); do echo $i; done
edit: I prefer seq over the other methods because I can actually remember it ;)
The seq method is the simplest, but Bash has built-in arithmetic evaluation.
END=5
for ((i=1;i<=END;i++)); do
echo $i
done
# ==> outputs 1 2 3 4 5 on separate lines
The for ((expr1;expr2;expr3)); construct works just like for (expr1;expr2;expr3) in C and similar languages, and like other ((expr)) cases, Bash treats them as arithmetic.
discussion
Using seq is fine, as Jiaaro suggested. Pax Diablo suggested a Bash loop to avoid calling a subprocess, with the additional advantage of being more memory friendly if $END is too large. Zathrus spotted a typical bug in the loop implementation, and also hinted that since i is a text variable, continuous conversions to-and-fro numbers are performed with an associated slow-down.
integer arithmetic
This is an improved version of the Bash loop:
typeset -i i END
let END=5 i=1
while ((i<=END)); do
echo $i
…
let i++
done
If the only thing that we want is the echo, then we could write echo $((i++)).
ephemient taught me something: Bash allows for ((expr;expr;expr)) constructs. Since I've never read the whole man page for Bash (like I've done with the Korn shell (ksh) man page, and that was a long time ago), I missed that.
So,
typeset -i i END # Let's be explicit
for ((i=1;i<=END;++i)); do echo $i; done
seems to be the most memory-efficient way (it won't be necessary to allocate memory to consume seq's output, which could be a problem if END is very large), although probably not the “fastest”.
the initial question
eschercycle noted that the {a..b} Bash notation works only with literals; true, accordingly to the Bash manual. One can overcome this obstacle with a single (internal) fork() without an exec() (as is the case with calling seq, which being another image requires a fork+exec):
for i in $(eval echo "{1..$END}"); do
Both eval and echo are Bash builtins, but a fork() is required for the command substitution (the $(…) construct).
Here is why the original expression didn't work.
From man bash:
Brace expansion is performed before
any other expansions, and any
characters special to other
expansions are preserved in the
result. It is strictly textual. Bash
does not apply any syntactic
interpretation to the context of
the expansion or the text between the
braces.
So, brace expansion is something done early as a purely textual macro operation, before parameter expansion.
Shells are highly optimized hybrids between macro processors and more formal programming languages. In order to optimize the typical use cases, the language is made rather more complex and some limitations are accepted.
Recommendation
I would suggest sticking with Posix1 features. This means using for i in <list>; do, if the list is already known, otherwise, use while or seq, as in:
#!/bin/sh
limit=4
i=1; while [ $i -le $limit ]; do
echo $i
i=$(($i + 1))
done
# Or -----------------------
for i in $(seq 1 $limit); do
echo $i
done
1. Bash is a great shell and I use it interactively, but I don't put bash-isms into my scripts. Scripts might need a faster shell, a more secure one, a more embedded-style one. They might need to run on whatever is installed as /bin/sh, and then there are all the usual pro-standards arguments. Remember shellshock, aka bashdoor?
The POSIX way
If you care about portability, use the example from the POSIX standard:
i=2
end=5
while [ $i -le $end ]; do
echo $i
i=$(($i+1))
done
Output:
2
3
4
5
Things which are not POSIX:
(( )) without dollar, although it is a common extension as mentioned by POSIX itself.
[[. [ is enough here. See also: What is the difference between single and double square brackets in Bash?
for ((;;))
seq (GNU Coreutils)
{start..end}, and that cannot work with variables as mentioned by the Bash manual.
let i=i+1: POSIX 7 2. Shell Command Language does not contain the word let, and it fails on bash --posix 4.3.42
the dollar at i=$i+1 might be required, but I'm not sure. POSIX 7 2.6.4 Arithmetic Expansion says:
If the shell variable x contains a value that forms a valid integer constant, optionally including a leading plus or minus sign, then the arithmetic expansions "$((x))" and "$(($x))" shall return the same value.
but reading it literally that does not imply that $((x+1)) expands since x+1 is not a variable.
You can use
for i in $(seq $END); do echo $i; done
Another layer of indirection:
for i in $(eval echo {1..$END}); do
∶
I've combined a few of the ideas here and measured performance.
TL;DR Takeaways:
seq and {..} are really fast
for and while loops are slow
$( ) is slow
for (( ; ; )) loops are slower
$(( )) is even slower
Worrying about N numbers in memory (seq or {..}) is silly (at least up to 1 million.)
These are not conclusions. You would have to look at the C code behind each of these to draw conclusions. This is more about how we tend to use each of these mechanisms for looping over code. Most single operations are close enough to being the same speed that it's not going to matter in most cases. But a mechanism like for (( i=1; i<=1000000; i++ )) is many operations as you can visually see. It is also many more operations per loop than you get from for i in $(seq 1 1000000). And that may not be obvious to you, which is why doing tests like this is valuable.
Demos
# show that seq is fast
$ time (seq 1 1000000 | wc)
1000000 1000000 6888894
real 0m0.227s
user 0m0.239s
sys 0m0.008s
# show that {..} is fast
$ time (echo {1..1000000} | wc)
1 1000000 6888896
real 0m1.778s
user 0m1.735s
sys 0m0.072s
# Show that for loops (even with a : noop) are slow
$ time (for i in {1..1000000} ; do :; done | wc)
0 0 0
real 0m3.642s
user 0m3.582s
sys 0m0.057s
# show that echo is slow
$ time (for i in {1..1000000} ; do echo $i; done | wc)
1000000 1000000 6888896
real 0m7.480s
user 0m6.803s
sys 0m2.580s
$ time (for i in $(seq 1 1000000) ; do echo $i; done | wc)
1000000 1000000 6888894
real 0m7.029s
user 0m6.335s
sys 0m2.666s
# show that C-style for loops are slower
$ time (for (( i=1; i<=1000000; i++ )) ; do echo $i; done | wc)
1000000 1000000 6888896
real 0m12.391s
user 0m11.069s
sys 0m3.437s
# show that arithmetic expansion is even slower
$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $i; i=$(($i+1)); done | wc)
1000000 1000000 6888896
real 0m19.696s
user 0m18.017s
sys 0m3.806s
$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $i; ((i=i+1)); done | wc)
1000000 1000000 6888896
real 0m18.629s
user 0m16.843s
sys 0m3.936s
$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $((i++)); done | wc)
1000000 1000000 6888896
real 0m17.012s
user 0m15.319s
sys 0m3.906s
# even a noop is slow
$ time (i=1; e=1000000; while [ $((i++)) -le $e ]; do :; done | wc)
0 0 0
real 0m12.679s
user 0m11.658s
sys 0m1.004s
If you need it prefix than you might like this
for ((i=7;i<=12;i++)); do echo `printf "%2.0d\n" $i |sed "s/ /0/"`;done
that will yield
07
08
09
10
11
12
If you're on BSD / OS X you can use jot instead of seq:
for i in $(jot $END); do echo $i; done
This works fine in bash:
END=5
i=1 ; while [[ $i -le $END ]] ; do
echo $i
((i = i + 1))
done
There are many ways to do this, however the ones I prefer is given below
Using seq
Synopsis from man seq
$ seq [-w] [-f format] [-s string] [-t string] [first [incr]] last
Syntax
Full command
seq first incr last
first is starting number in the sequence [is optional, by default:1]
incr is increment [is optional, by default:1]
last is the last number in the sequence
Example:
$ seq 1 2 10
1 3 5 7 9
Only with first and last:
$ seq 1 5
1 2 3 4 5
Only with last:
$ seq 5
1 2 3 4 5
Using {first..last..incr}
Here first and last are mandatory and incr is optional
Using just first and last
$ echo {1..5}
1 2 3 4 5
Using incr
$ echo {1..10..2}
1 3 5 7 9
You can use this even for characters like below
$ echo {a..z}
a b c d e f g h i j k l m n o p q r s t u v w x y z
I know this question is about bash, but - just for the record - ksh93 is smarter and implements it as expected:
$ ksh -c 'i=5; for x in {1..$i}; do echo "$x"; done'
1
2
3
4
5
$ ksh -c 'echo $KSH_VERSION'
Version JM 93u+ 2012-02-29
$ bash -c 'i=5; for x in {1..$i}; do echo "$x"; done'
{1..5}
This is another way:
end=5
for i in $(bash -c "echo {1..${end}}"); do echo $i; done
If you want to stay as close as possible to the brace-expression syntax, try out the range function from bash-tricks' range.bash.
For example, all of the following will do the exact same thing as echo {1..10}:
source range.bash
one=1
ten=10
range {$one..$ten}
range $one $ten
range {1..$ten}
range {1..10}
It tries to support the native bash syntax with as few "gotchas" as possible: not only are variables supported, but the often-undesirable behavior of invalid ranges being supplied as strings (e.g. for i in {1..a}; do echo $i; done) is prevented as well.
The other answers will work in most cases, but they all have at least one of the following drawbacks:
Many of them use subshells, which can harm performance and may not be possible on some systems.
Many of them rely on external programs. Even seq is a binary which must be installed to be used, must be loaded by bash, and must contain the program you expect, for it to work in this case. Ubiquitous or not, that's a lot more to rely on than just the Bash language itself.
Solutions that do use only native Bash functionality, like #ephemient's, will not work on alphabetic ranges, like {a..z}; brace expansion will. The question was about ranges of numbers, though, so this is a quibble.
Most of them aren't visually similar to the {1..10} brace-expanded range syntax, so programs that use both may be a tiny bit harder to read.
#bobbogo's answer uses some of the familiar syntax, but does something unexpected if the $END variable is not a valid range "bookend" for the other side of the range. If END=a, for example, an error will not occur and the verbatim value {1..a} will be echoed. This is the default behavior of Bash, as well--it is just often unexpected.
Disclaimer: I am the author of the linked code.
These are all nice but seq is supposedly deprecated and most only work with numeric ranges.
If you enclose your for loop in double quotes, the start and end variables will be dereferenced when you echo the string, and you can ship the string right back to BASH for execution. $i needs to be escaped with \'s so it is NOT evaluated before being sent to the subshell.
RANGE_START=a
RANGE_END=z
echo -e "for i in {$RANGE_START..$RANGE_END}; do echo \\${i}; done" | bash
This output can also be assigned to a variable:
VAR=`echo -e "for i in {$RANGE_START..$RANGE_END}; do echo \\${i}; done" | bash`
The only "overhead" this should generate should be the second instance of bash so it should be suitable for intensive operations.
Replace {} with (( )):
tmpstart=0;
tmpend=4;
for (( i=$tmpstart; i<=$tmpend; i++ )) ; do
echo $i ;
done
Yields:
0
1
2
3
4
If you're doing shell commands and you (like I) have a fetish for pipelining, this one is good:
seq 1 $END | xargs -I {} echo {}
if you don't wanna use 'seq' or 'eval' or jot or arithmetic expansion format eg. for ((i=1;i<=END;i++)), or other loops eg. while, and you don't wanna 'printf' and happy to 'echo' only, then this simple workaround might fit your budget:
a=1; b=5; d='for i in {'$a'..'$b'}; do echo -n "$i"; done;' echo "$d" | bash
PS: My bash doesn't have 'seq' command anyway.
Tested on Mac OSX 10.6.8, Bash 3.2.48
This works in Bash and Korn, also can go from higher to lower numbers. Probably not fastest or prettiest but works well enough. Handles negatives too.
function num_range {
# Return a range of whole numbers from beginning value to ending value.
# >>> num_range start end
# start: Whole number to start with.
# end: Whole number to end with.
typeset s e v
s=${1}
e=${2}
if (( ${e} >= ${s} )); then
v=${s}
while (( ${v} <= ${e} )); do
echo ${v}
((v=v+1))
done
elif (( ${e} < ${s} )); then
v=${s}
while (( ${v} >= ${e} )); do
echo ${v}
((v=v-1))
done
fi
}
function test_num_range {
num_range 1 3 | egrep "1|2|3" | assert_lc 3
num_range 1 3 | head -1 | assert_eq 1
num_range -1 1 | head -1 | assert_eq "-1"
num_range 3 1 | egrep "1|2|3" | assert_lc 3
num_range 3 1 | head -1 | assert_eq 3
num_range 1 -1 | tail -1 | assert_eq "-1"
}
Related
I wrote small script under Debian Linux 11 that should check how many instances of application is currently running and what is power usage of GPU cards.
I save it under name test , and she is started every time I access instance over SSH
#!/bin/sh
clear
a=$(nvidia-smi -q -i 0 | grep "Power Draw" | cut -c45-50)
b=$(nvidia-smi -q -i 1 | grep "Power Draw" | cut -c45-50)
c=$(nvidia-smi -q -i 2 | grep "Power Draw" | cut -c45-50)
d=$(nvidia-smi -q -i 3 | grep "Power Draw" | cut -c45-50)
zet=$( echo "$a" + "$b" + "$c" + "$d" | bc -l )
echo "SYSTEM DRAW:" "$zet"
if [ "${zet}" -gt 150 ]; then
echo WARRNING - SYSTEM DRAW LOW
else
echo OK
fi
sleep 8
exit
All I need to add is this line
x=${x%.*}
That convert decimal number in number without decimals and script works perfect.
You could add set -x right before the part you want to debug, which will show you a debug of what is happening in bash. and stop it by inserting after that set +x
like:
set -x
power=$150
echo "SYSTEM DRAW :" $total
if [ $total \> $power ] ; then # escape > otherwise it redirects output
I do not think you are setting the value of $150
The script might be failing if one of the compared values is not set.. so you should initialize your variables to be let' say equal to 0 as a default at the beginning of your script, or via bash
like:
power=${150:-10} # if $150 does not have a value or empty, the default value of $power will be set to `10`
So many possible issues but you can compare possibly decimal values using bc:
if [ "$(echo "$total > $power" | bc)" = 1 ]; then
One problem is that [ (and [[) do string comparisons, and you want a numeric comparison. For that you want to use ((, so something like
if (( $total > 150 )); then
echo "WARNING..."
else
echo "OK"
fi
will work better. Otherwise a total of 1000 would print ok, and 90 would give a warning.
Other problems:
$150 gives you the value of a variable called 150 -- you probably want to remove the $
Outside of special forms like [[ and ((, a > will do an output redirection, rather than being a 'normal' argument to a command such as [.
As the comments recommend, use shellcheck, however, I think your intention is not what you wrote.
Try this, create a script (i.e. myscripy)
#! /bin/bash
power=$150
echo "power=$power"
then run it
./myscript abc
and prints
power=abc50
which is probably very different than what you expect.
That is because power will take the first argument's value ($1) and append 50.
If you wanted argument number 150 (also very unlikely), you should write
power=${150}
but if you want just the number
power=150
edit
based on the comment, use
zet=$(bc <<<"$a+$b+$c+$d")
to calculate zet if the values are floating points.
For the comparison use
if [ "$(bc <<<"$zet>150")" == 1 ]; then
...
fi
I have this Bash script and I had a problem in line 16.
How can I take the previous result of line 15 and add
it to the variable in line 16?
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in `ls output-$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"
How do I iterate over a range of numbers in Bash when the range is given by a variable?
I know I can do this (called "sequence expression" in the Bash documentation):
for i in {1..5}; do echo $i; done
Which gives:
1
2
3
4
5
Yet, how can I replace either of the range endpoints with a variable? This doesn't work:
END=5
for i in {1..$END}; do echo $i; done
Which prints:
{1..5}
for i in $(seq 1 $END); do echo $i; done
edit: I prefer seq over the other methods because I can actually remember it ;)
The seq method is the simplest, but Bash has built-in arithmetic evaluation.
END=5
for ((i=1;i<=END;i++)); do
echo $i
done
# ==> outputs 1 2 3 4 5 on separate lines
The for ((expr1;expr2;expr3)); construct works just like for (expr1;expr2;expr3) in C and similar languages, and like other ((expr)) cases, Bash treats them as arithmetic.
discussion
Using seq is fine, as Jiaaro suggested. Pax Diablo suggested a Bash loop to avoid calling a subprocess, with the additional advantage of being more memory friendly if $END is too large. Zathrus spotted a typical bug in the loop implementation, and also hinted that since i is a text variable, continuous conversions to-and-fro numbers are performed with an associated slow-down.
integer arithmetic
This is an improved version of the Bash loop:
typeset -i i END
let END=5 i=1
while ((i<=END)); do
echo $i
…
let i++
done
If the only thing that we want is the echo, then we could write echo $((i++)).
ephemient taught me something: Bash allows for ((expr;expr;expr)) constructs. Since I've never read the whole man page for Bash (like I've done with the Korn shell (ksh) man page, and that was a long time ago), I missed that.
So,
typeset -i i END # Let's be explicit
for ((i=1;i<=END;++i)); do echo $i; done
seems to be the most memory-efficient way (it won't be necessary to allocate memory to consume seq's output, which could be a problem if END is very large), although probably not the “fastest”.
the initial question
eschercycle noted that the {a..b} Bash notation works only with literals; true, accordingly to the Bash manual. One can overcome this obstacle with a single (internal) fork() without an exec() (as is the case with calling seq, which being another image requires a fork+exec):
for i in $(eval echo "{1..$END}"); do
Both eval and echo are Bash builtins, but a fork() is required for the command substitution (the $(…) construct).
Here is why the original expression didn't work.
From man bash:
Brace expansion is performed before
any other expansions, and any
characters special to other
expansions are preserved in the
result. It is strictly textual. Bash
does not apply any syntactic
interpretation to the context of
the expansion or the text between the
braces.
So, brace expansion is something done early as a purely textual macro operation, before parameter expansion.
Shells are highly optimized hybrids between macro processors and more formal programming languages. In order to optimize the typical use cases, the language is made rather more complex and some limitations are accepted.
Recommendation
I would suggest sticking with Posix1 features. This means using for i in <list>; do, if the list is already known, otherwise, use while or seq, as in:
#!/bin/sh
limit=4
i=1; while [ $i -le $limit ]; do
echo $i
i=$(($i + 1))
done
# Or -----------------------
for i in $(seq 1 $limit); do
echo $i
done
1. Bash is a great shell and I use it interactively, but I don't put bash-isms into my scripts. Scripts might need a faster shell, a more secure one, a more embedded-style one. They might need to run on whatever is installed as /bin/sh, and then there are all the usual pro-standards arguments. Remember shellshock, aka bashdoor?
The POSIX way
If you care about portability, use the example from the POSIX standard:
i=2
end=5
while [ $i -le $end ]; do
echo $i
i=$(($i+1))
done
Output:
2
3
4
5
Things which are not POSIX:
(( )) without dollar, although it is a common extension as mentioned by POSIX itself.
[[. [ is enough here. See also: What is the difference between single and double square brackets in Bash?
for ((;;))
seq (GNU Coreutils)
{start..end}, and that cannot work with variables as mentioned by the Bash manual.
let i=i+1: POSIX 7 2. Shell Command Language does not contain the word let, and it fails on bash --posix 4.3.42
the dollar at i=$i+1 might be required, but I'm not sure. POSIX 7 2.6.4 Arithmetic Expansion says:
If the shell variable x contains a value that forms a valid integer constant, optionally including a leading plus or minus sign, then the arithmetic expansions "$((x))" and "$(($x))" shall return the same value.
but reading it literally that does not imply that $((x+1)) expands since x+1 is not a variable.
You can use
for i in $(seq $END); do echo $i; done
Another layer of indirection:
for i in $(eval echo {1..$END}); do
∶
I've combined a few of the ideas here and measured performance.
TL;DR Takeaways:
seq and {..} are really fast
for and while loops are slow
$( ) is slow
for (( ; ; )) loops are slower
$(( )) is even slower
Worrying about N numbers in memory (seq or {..}) is silly (at least up to 1 million.)
These are not conclusions. You would have to look at the C code behind each of these to draw conclusions. This is more about how we tend to use each of these mechanisms for looping over code. Most single operations are close enough to being the same speed that it's not going to matter in most cases. But a mechanism like for (( i=1; i<=1000000; i++ )) is many operations as you can visually see. It is also many more operations per loop than you get from for i in $(seq 1 1000000). And that may not be obvious to you, which is why doing tests like this is valuable.
Demos
# show that seq is fast
$ time (seq 1 1000000 | wc)
1000000 1000000 6888894
real 0m0.227s
user 0m0.239s
sys 0m0.008s
# show that {..} is fast
$ time (echo {1..1000000} | wc)
1 1000000 6888896
real 0m1.778s
user 0m1.735s
sys 0m0.072s
# Show that for loops (even with a : noop) are slow
$ time (for i in {1..1000000} ; do :; done | wc)
0 0 0
real 0m3.642s
user 0m3.582s
sys 0m0.057s
# show that echo is slow
$ time (for i in {1..1000000} ; do echo $i; done | wc)
1000000 1000000 6888896
real 0m7.480s
user 0m6.803s
sys 0m2.580s
$ time (for i in $(seq 1 1000000) ; do echo $i; done | wc)
1000000 1000000 6888894
real 0m7.029s
user 0m6.335s
sys 0m2.666s
# show that C-style for loops are slower
$ time (for (( i=1; i<=1000000; i++ )) ; do echo $i; done | wc)
1000000 1000000 6888896
real 0m12.391s
user 0m11.069s
sys 0m3.437s
# show that arithmetic expansion is even slower
$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $i; i=$(($i+1)); done | wc)
1000000 1000000 6888896
real 0m19.696s
user 0m18.017s
sys 0m3.806s
$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $i; ((i=i+1)); done | wc)
1000000 1000000 6888896
real 0m18.629s
user 0m16.843s
sys 0m3.936s
$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $((i++)); done | wc)
1000000 1000000 6888896
real 0m17.012s
user 0m15.319s
sys 0m3.906s
# even a noop is slow
$ time (i=1; e=1000000; while [ $((i++)) -le $e ]; do :; done | wc)
0 0 0
real 0m12.679s
user 0m11.658s
sys 0m1.004s
If you need it prefix than you might like this
for ((i=7;i<=12;i++)); do echo `printf "%2.0d\n" $i |sed "s/ /0/"`;done
that will yield
07
08
09
10
11
12
If you're on BSD / OS X you can use jot instead of seq:
for i in $(jot $END); do echo $i; done
This works fine in bash:
END=5
i=1 ; while [[ $i -le $END ]] ; do
echo $i
((i = i + 1))
done
There are many ways to do this, however the ones I prefer is given below
Using seq
Synopsis from man seq
$ seq [-w] [-f format] [-s string] [-t string] [first [incr]] last
Syntax
Full command
seq first incr last
first is starting number in the sequence [is optional, by default:1]
incr is increment [is optional, by default:1]
last is the last number in the sequence
Example:
$ seq 1 2 10
1 3 5 7 9
Only with first and last:
$ seq 1 5
1 2 3 4 5
Only with last:
$ seq 5
1 2 3 4 5
Using {first..last..incr}
Here first and last are mandatory and incr is optional
Using just first and last
$ echo {1..5}
1 2 3 4 5
Using incr
$ echo {1..10..2}
1 3 5 7 9
You can use this even for characters like below
$ echo {a..z}
a b c d e f g h i j k l m n o p q r s t u v w x y z
I know this question is about bash, but - just for the record - ksh93 is smarter and implements it as expected:
$ ksh -c 'i=5; for x in {1..$i}; do echo "$x"; done'
1
2
3
4
5
$ ksh -c 'echo $KSH_VERSION'
Version JM 93u+ 2012-02-29
$ bash -c 'i=5; for x in {1..$i}; do echo "$x"; done'
{1..5}
This is another way:
end=5
for i in $(bash -c "echo {1..${end}}"); do echo $i; done
If you want to stay as close as possible to the brace-expression syntax, try out the range function from bash-tricks' range.bash.
For example, all of the following will do the exact same thing as echo {1..10}:
source range.bash
one=1
ten=10
range {$one..$ten}
range $one $ten
range {1..$ten}
range {1..10}
It tries to support the native bash syntax with as few "gotchas" as possible: not only are variables supported, but the often-undesirable behavior of invalid ranges being supplied as strings (e.g. for i in {1..a}; do echo $i; done) is prevented as well.
The other answers will work in most cases, but they all have at least one of the following drawbacks:
Many of them use subshells, which can harm performance and may not be possible on some systems.
Many of them rely on external programs. Even seq is a binary which must be installed to be used, must be loaded by bash, and must contain the program you expect, for it to work in this case. Ubiquitous or not, that's a lot more to rely on than just the Bash language itself.
Solutions that do use only native Bash functionality, like #ephemient's, will not work on alphabetic ranges, like {a..z}; brace expansion will. The question was about ranges of numbers, though, so this is a quibble.
Most of them aren't visually similar to the {1..10} brace-expanded range syntax, so programs that use both may be a tiny bit harder to read.
#bobbogo's answer uses some of the familiar syntax, but does something unexpected if the $END variable is not a valid range "bookend" for the other side of the range. If END=a, for example, an error will not occur and the verbatim value {1..a} will be echoed. This is the default behavior of Bash, as well--it is just often unexpected.
Disclaimer: I am the author of the linked code.
These are all nice but seq is supposedly deprecated and most only work with numeric ranges.
If you enclose your for loop in double quotes, the start and end variables will be dereferenced when you echo the string, and you can ship the string right back to BASH for execution. $i needs to be escaped with \'s so it is NOT evaluated before being sent to the subshell.
RANGE_START=a
RANGE_END=z
echo -e "for i in {$RANGE_START..$RANGE_END}; do echo \\${i}; done" | bash
This output can also be assigned to a variable:
VAR=`echo -e "for i in {$RANGE_START..$RANGE_END}; do echo \\${i}; done" | bash`
The only "overhead" this should generate should be the second instance of bash so it should be suitable for intensive operations.
Replace {} with (( )):
tmpstart=0;
tmpend=4;
for (( i=$tmpstart; i<=$tmpend; i++ )) ; do
echo $i ;
done
Yields:
0
1
2
3
4
If you're doing shell commands and you (like I) have a fetish for pipelining, this one is good:
seq 1 $END | xargs -I {} echo {}
if you don't wanna use 'seq' or 'eval' or jot or arithmetic expansion format eg. for ((i=1;i<=END;i++)), or other loops eg. while, and you don't wanna 'printf' and happy to 'echo' only, then this simple workaround might fit your budget:
a=1; b=5; d='for i in {'$a'..'$b'}; do echo -n "$i"; done;' echo "$d" | bash
PS: My bash doesn't have 'seq' command anyway.
Tested on Mac OSX 10.6.8, Bash 3.2.48
This works in Bash and Korn, also can go from higher to lower numbers. Probably not fastest or prettiest but works well enough. Handles negatives too.
function num_range {
# Return a range of whole numbers from beginning value to ending value.
# >>> num_range start end
# start: Whole number to start with.
# end: Whole number to end with.
typeset s e v
s=${1}
e=${2}
if (( ${e} >= ${s} )); then
v=${s}
while (( ${v} <= ${e} )); do
echo ${v}
((v=v+1))
done
elif (( ${e} < ${s} )); then
v=${s}
while (( ${v} >= ${e} )); do
echo ${v}
((v=v-1))
done
fi
}
function test_num_range {
num_range 1 3 | egrep "1|2|3" | assert_lc 3
num_range 1 3 | head -1 | assert_eq 1
num_range -1 1 | head -1 | assert_eq "-1"
num_range 3 1 | egrep "1|2|3" | assert_lc 3
num_range 3 1 | head -1 | assert_eq 3
num_range 1 -1 | tail -1 | assert_eq "-1"
}
I have the following function to count the number of differences between two Strings:
get_num_diffs_strict() {
local -i diffs=$(cmp -l <(echo "$1") <(echo "$2") 2>/dev/null | wc -l)
local -i lendiff=0
if (( ${#1} > ${#2} ));then
lendiff=$((${#1} - ${#2}))
else
lendiff=$((${#2} - ${#1}))
fi
if (( lendiff > 0 ));then # lines differ in length
(( lendiff-- )) # subtract already accounted EOF character
(( diffs+=lendiff ))
fi
echo $diffs
}
For inputs like
$ get_num_diffs_strict hello somethingelse
$ 13
$ get_num_diffs_strict hello holla
$ 2
the results are acceptable.
For other input strings I would like it to be less strict with what is deemed as a difference. Consider the following cases
$ get_num_diffs_strict hello helo
$ 2
$ get_num_diffs_strict hello heello
$ 3
$ get_num_diffs_strict 0123456789 0123567889
$ 4
The desired output, however, should be 1 in the first two cases and 2 in the last one.
In case 1 and 2 there is only one letter missing or just one additional letter. It should count as one difference. In the last case the second string is missing the number 4 and contains the number 8 twice. These should count only as two differences - like a human would intuitively perceive it.
Is there any program, similar to cmp, that can accomplish that? Alternative solutions (python, perl, etc.) are also acceptable. I am merely trying to avoid reinventing the wheel.
Here is my simple shell code. I want the result to be 2.Shell treats everything as a string.
How can i get this done ?
num=1
num=$(( $num + 1 ))
EDIT :
Complete code : Whats wrong in this if i want to print from 1 to 10 ?
#! /bin/bash
num=1
until test $num -eq 10
do
num=$(( $num + 1 ))
echo $num
done
In bash, you don't need to do anything special:
aix#aix:~$ num=1
aix#aix:~$ num=$(( $num + 1 ))
aix#aix:~$ echo $num
2
#tonio; please don't advocate using subshell (` ... or $( ... ) ) constructs when they're not needed (to keep confusion to the maximum, $(( ... )) is not a sub-shell construct). Sub-shells can make a tremendous performance hit even with rather trivial amounts of data. The same is true for every place where an external program is used to do somethign that could be done with a shel built-in.
Example:
num=1
time while [[ $num -lt 10000 ]]; do
num=$(( num+1 ))
done
echo $num
num=1
time while /bin/test $num -lt 10000; do
num=$( /bin/expr $num + 1 )
done
echo $num
Output (run in ksh on Linux):
real 0m0.04s
user 0m0.04s
sys 0m0.01s
10000
real 0m20.32s
user 0m2.23s
sys 0m2.92s
10000
...so run-time factor of 250, and CPU-time factor of 100. I admit the example I used was a exaggerated one, with explicitly requiring all built-ins to be bypassed, but I think the point was made: creating new processes is expenisve, avoid it when you can, and know your shell to recognise where new processes are created.
This might work for you:
num=1; ((num++)); echo $num
2
or
num=1; echo $((++num))
2
for loops
for num in {1..10}; do echo $num; done
or (in bash at least)
for ((num=1; num<=10; num++)) { echo $num; }
second loop more useful when more programming involved:
for (( num=1,mun=10; num<=10; num++,mun--)) { echo $num $mun; }
You just did:
$ num=1; num=$(( $num + 1 ));echo $num
2
Note: You don't need to quote variables inside $(( )). Also, you can just use $((num++))
You are not specifying which shell you are using, but the most concise form I know is this one (works at least in bash):
num=$[num+1]
If only incrementing by one and changing the variable itself rather than printing/assigning, then:
((num++))
Is a better/more elegant solution. See dogbane's answer for that.
If looping over the values, I would use this form instead:
for i in `seq 1 10`; do
echo $i
done
Use ((num++)) as shorthand for incrementing num.
$ num=1
$ ((num++))
$ echo $num
2
try this
$ num=1; num=`expr $num + 1`; echo $num;
EDIT:
More efficient would be:
num=$(( num + 1 ))
Thanks #Charles Duffy for your comment.
works for me
$ num=1
$ num=$(( $num + 1 ))
$ echo $num
2
What output do you get?
Read more about bash Arithmetic # tldp
EDIT
To do something 10 times in bash you can use (using brace-expansion}
$ for i in {1..10}; do echo $i; done
1
2
3
4
5
6
7
8
9
10
However, you cannot use variables between the {}. If this is the case, use seq instead.