echo 3+3
How can I evaluate such expressions in Bash, in this case to 6?
echo $(( 3+3 ))
expr is the standard way, but it only handles integers.
bash has a couple of extensions, which only handle integers as well:
$((3+3)) returns 6
((3+3)) used in conditionals, returns 0 for true (non-zero) and 1 for false
let 3+3 same as (( ))
let and (( )) can be used to assign values, e.g.
let a=3+3
((a=3+3))
for floating point you can use bc
echo 3+3 | bc
in shells such as zsh/ksh, you can use floats for maths. If you need more maths power, use tools like bc/awk/dc
eg
var=$(echo "scale=2;3.4+43.1" | bc)
var=$(awk 'BEGIN{print 3.4*43.1}')
looking at what you are trying to do
awk '{printf "%.2f\n",$0/59.5}' ball_dropping_times >bull_velocities
You can make use of the expr command as:
expr 3 + 3
To store the result into a variable you can do:
sum=$(expr 3 + 3)
or
sum=`expr 3 + 3`
Lots of ways - most portable is to use the expr command:
expr 3 + 3
I believe the ((3+3)) method is the most rapid as it's interpreted by the shell rather than an external binary.
time a large loop using all suggested methods for the most efficient.
Solved thanks to Dennis, an example of BC-use:
$ cat calc_velo.sh
#!/bin/bash
for i in `cat ball_dropping_times`
do
echo "scale=20; $i / 59.5" | bc
done > ball_velocities
My understanding of math processing involves floating point processing.
Using bashj (https://sourceforge.net/projects/bashj/) you can call a java method (with floating point processing, cos(), sin(), log(), exp()...) using simply
bashj +eval "3+3"
bashj +eval "3.5*5.5"
or in a bashj script, java calls of this kind:
#!/usr/bin/bashj
EXPR="3.0*6.0"
echo $EXPR "=" u.doubleEval($EXPR)
FUNCTIONX="3*x*x+cos(x)+1"
X=3.0
FX=u.doubleEval($FUNCTIONX,$X)
echo "x="$X " => f(x)=" $FUNCTIONX "=" $FX
Note the interesting speed : ~ 10 msec per call (the answer is provided by a JVM server).
Note also that u.doubleEval(1/2) will provide 0.5 (floating point) instead of 0 (integer)
One use case that might be useful in this regard is, if one of your operand itself is a bash command then try this.
echo $(( `date +%s\`+10 )) or even echo $(( `date +%s\`+(60*60) ))
In my case I was trying to get Unixtime 10 seconds and hour later than current time respectively.
Related
echo 3+3
How can I evaluate such expressions in Bash, in this case to 6?
echo $(( 3+3 ))
expr is the standard way, but it only handles integers.
bash has a couple of extensions, which only handle integers as well:
$((3+3)) returns 6
((3+3)) used in conditionals, returns 0 for true (non-zero) and 1 for false
let 3+3 same as (( ))
let and (( )) can be used to assign values, e.g.
let a=3+3
((a=3+3))
for floating point you can use bc
echo 3+3 | bc
in shells such as zsh/ksh, you can use floats for maths. If you need more maths power, use tools like bc/awk/dc
eg
var=$(echo "scale=2;3.4+43.1" | bc)
var=$(awk 'BEGIN{print 3.4*43.1}')
looking at what you are trying to do
awk '{printf "%.2f\n",$0/59.5}' ball_dropping_times >bull_velocities
You can make use of the expr command as:
expr 3 + 3
To store the result into a variable you can do:
sum=$(expr 3 + 3)
or
sum=`expr 3 + 3`
Lots of ways - most portable is to use the expr command:
expr 3 + 3
I believe the ((3+3)) method is the most rapid as it's interpreted by the shell rather than an external binary.
time a large loop using all suggested methods for the most efficient.
Solved thanks to Dennis, an example of BC-use:
$ cat calc_velo.sh
#!/bin/bash
for i in `cat ball_dropping_times`
do
echo "scale=20; $i / 59.5" | bc
done > ball_velocities
My understanding of math processing involves floating point processing.
Using bashj (https://sourceforge.net/projects/bashj/) you can call a java method (with floating point processing, cos(), sin(), log(), exp()...) using simply
bashj +eval "3+3"
bashj +eval "3.5*5.5"
or in a bashj script, java calls of this kind:
#!/usr/bin/bashj
EXPR="3.0*6.0"
echo $EXPR "=" u.doubleEval($EXPR)
FUNCTIONX="3*x*x+cos(x)+1"
X=3.0
FX=u.doubleEval($FUNCTIONX,$X)
echo "x="$X " => f(x)=" $FUNCTIONX "=" $FX
Note the interesting speed : ~ 10 msec per call (the answer is provided by a JVM server).
Note also that u.doubleEval(1/2) will provide 0.5 (floating point) instead of 0 (integer)
One use case that might be useful in this regard is, if one of your operand itself is a bash command then try this.
echo $(( `date +%s\`+10 )) or even echo $(( `date +%s\`+(60*60) ))
In my case I was trying to get Unixtime 10 seconds and hour later than current time respectively.
I would like to compare a number, which is the output of a command, with a constant and do some manipulation. That is, if $id < 10, I want to see 590$id and if it is above 10, I want to see 59$id.
I found that expr doesn't working here:
ID=3
NUM=59$ID
if [ `expr $ID` -lt 10]; then
NUM=590$ID
fi
echo $NUM
The output of the code is 593 and not 5903. Even, $(($ID + 5900)) -lt 5910 writes 593.
How can I fix that?
Could you please try following.
cat script.sh
#!/bin/bash
ID=$(printf "%02d" 3 )
##NUM=59$ID ##Commented this to check if, condition is getting satisfied or not. Doesn't seem to be fit here.
(( $ID < 10 )) && NUM="59$ID"
echo "$NUM"
Output will be 5903 after running above code.
Don't use expr. It's old and tricky.
Don't use backticks `. They are discouraged and $( ... ) is preferred.
For arithmetic comparisons use arithmetic expansions. Just
if (( ID < 10 )); then
Note that bash is space aware and your script has a syntax error, it is missing a space - the 10]; should be 10 ];.
Note that by convention uppercase variables should be used for exported variables.
Looking at your code I think you just want:
NUM=$((5900 + ID))
I want to use a string to control a for loop in bash. My first test code produces what I would expect and what I want:
$ aa='1 2 3 4'
$ for ii in $aa; do echo $ii; done
1
2
3
4
I'd like to use something like the following instead. This doesn't give the output I'd like (I can see why it does what it does).
$ aa='1..4'
$ for ii in $aa; do echo $ii; done
1..4
Any suggestions on how I should modify the second example to give the same output as the first?
Thanks in advance for any thoughts. I'm slowly learning bash but still have a lot to learn.
Mike
The notation could be written out as:
for ii in {1..4}; do echo "$ii"; done
but the {1..4} needs to be written out like that, no variables involved, and not as the result of variable substitution. That is brace expansion in the Bash manual, and it happens before string expansions, etc. You'll probably be best off using:
for ii in $(seq 1 4); do echo "$ii"; done
where either the 1 or the 4 or both can be shell variables.
You could use seq command (see man seq).
$ aa='1 4'
$ for ii in $(seq $aa); do echo $ii; done
Bash won't do brace expansion with variables, but you can use eval:
$ aa='1..4'
$ for ii in $(eval echo {$aa}); do echo $ii; done
1
2
3
4
You could also split aa into an array:
IFS=. arr=($aa)
for ((ii=arr[0]; ii<arr[2]; ii++)); do echo $ii; done
Note that IFS can only be a single character, so the .. range places the numbers into indexes 0 and 2.
Note There are certainly more elegant ways of doing this, as Ben Grimm's answer, and this is not pure bash, as uses seq and awk.
One way of achieving this is by calling seq. It would be trivial if you knew the numbers in the string beforehand, so there would be no need to do any conversion, as you could simple do seq 1 4 or seq $a $b for that matter.
I assume, however, that your input is indeed a string in the format you mentioned, that is, 1..4 or 20..100. For this purpose you could convert the string into 2 numbers ans use them as parameters for seq.
One of possibly many ways of achieving this is:
$ `echo "1..4" | sed -e 's/\.\./ /g' | awk '{print "seq", $1, $2}'`
1
2
3
4
Note that this will work the same way for any input in the given format. If desired, sed can be changed by tr with similar results.
$ x="10..15"
$ `echo $x | tr "." " " | awk '{print "seq", $1, $2}'`
10
11
12
13
14
15
I am new to bash scripting. Now, the question is self explanatory. I want to print the values 2^0, 2^1, 2^2, 2^3, 2^4, 2^5 using loop in bash.
I tried ..
for i in {0...5}; do echo 2^$i; done
result:
2^{0...5}
please suggest a solution
This is the correct form:
for i in {0..5}; do echo $((2**i)); done
Where { .. } is the range operator and $(( )) the arithmetical evaluation operator.
Note that the power operator in Bash is written ** and not ^.
Edit: Roberto Reale's answer is much better for this purpose, as it's shorter and simpler.
2^$i; won't work unfortunately, as bash doesn't support that operator.
You could try adding this
pow()
{
echo $(( ${1:?} ** ${2:?} ))
}
to your code (using it like pow [base] [exponent] (without the brackets of course)).
(Bare in mind I havent't tested this, so this may not work as expected)
Source
You could try this code also,
$ for i in {0..5}; do awk -v var=$i 'BEGIN{print 2^var}'; done
1
2
4
8
16
32
In a text file, test.txt, I have the next information:
sl-gs5 desconnected Wed Oct 10 08:00:01 EDT 2012 1001
I want to extract the hour of the event by the next command line:
hour=$(grep -n sl-gs5 test.txt | tail -1 | cut -d' ' -f6 | awk -F ":" '{print $1}')
and I got "08". When I try to add 1,
14 echo $((hour+1))
I receive the next error message:
./test2.sh: line 14: 08: value too great for base (error token is "08")
If variables in Bash are untyped, why?
See ARITHMETIC EVALUATION in man bash:
Constants with a leading 0 are interpreted as octal numbers.
You can remove the leading zero by parameter expansion:
hour=${hour#0}
or force base-10 interpretation:
$((10#$hour + 1))
what I'd call a hack, but given that you're only processing hour values, you can do
hour=08
echo $(( ${hour#0} +1 ))
9
hour=10
echo $(( ${hour#0} +1))
11
with little risk.
IHTH.
You could also use bc
hour=8
result=$(echo "$hour + 1" | bc)
echo $result
9
Here's an easy way, albeit not the prettiest way to get an int value for a string.
hour=`expr $hour + 0`
Example
bash-3.2$ hour="08"
bash-3.2$ hour=`expr $hour + 0`
bash-3.2$ echo $hour
8
In Short: In order to deal with "Leading Zero" numbers (any 0 digit that comes before the first non-zero) in bash
- Use bc An arbitrary precision calculator language
Example:
a="000001"
b=$(echo $a | bc)
echo $b
Output: 1
From Bash manual:
"bc is a language that supports arbitrary precision numbers with interactive execution
of statements. There are some similarities in the syntax to the C programming lan-
guage. A standard math library is available by command line option. If requested, the
math library is defined before processing any files. bc starts by processing code from
all the files listed on the command line in the order listed. After all files have
been processed, bc reads from the standard input. All code is executed as it is read.
(If a file contains a command to halt the processor, bc will never read from the standard input.)"
Since hours are always positive, and always 2 digits, you can set a 1 in front of it and subtract 100:
echo $((1$hour+1-100))
which is equivalent to
echo $((1$hour-99))
Be sure to comment such gymnastics. :)
The leading 0 is leading to bash trying to interpret your number as an octal number, but octal numbers are 0-7, and 8 is thus an invalid token.
If I were you, I would add some logic to remove a leading 0, add one, and re-add the leading 0 if the result is < 10.
How about sed?
hour=`echo $hour|sed -e "s/^0*//g"`