Bash float comparison error with different number of digits [duplicate] - bash

This question already has answers here:
How can I compare two floating point numbers in Bash?
(22 answers)
Closed 4 years ago.
Ok, so I've been working around with some results from speedtest-cli and realised that I had some errors due to the fact that bash doesn't seem to correctly handle the change in digits?
Anyways, here is an example ran directly from the terminal :
ubuntu:~$ l1=9.99
ubuntu:~$ l2=10.44
ubuntu:~$ if [[ (($l2 > $l1)) ]]; then echo "Ok"; fi
ubuntu:~$ if [[ (($l2 < $l1)) ]]; then echo "Not ok"; fi
Not ok
Of course, comparing eg. 10.33 and 11.34 would give the right result.
How does this happen and how can I fix it? Is there another way to achieve this comparison?
Thanks

You're using string comparison, not numeric. Inside double square brackets, parentheses are used just for precedence, so your condition is equivalent to
[[ $l2 < $l1 ]]
To use numeric comparison, use double parentheses without the square ones:
(( l2 < l1 ))
Unfortunately, this wouldn't work either, as bash doesn't support floating point arithmetic, only integer.
You need to use an external tool, e.g.
bc <<< "$l1 < $l2"
bc returns 1 for true and 0 for false.

Related

Comparisons in Shell [duplicate]

This question already has answers here:
Why should there be spaces around '[' and ']' in Bash?
(5 answers)
Closed 4 years ago.
I am trying to write a program that checks whether the number stored in a file (variable n) + 8 is greater or equal to 100. If it is, terminate, else, add 8 and store back in file. However, when I try running it, it says the command in line 4 (if condition) cannot be found. Can someone please explain to me why this isn't working? Thanks.
#!/bin/bash
n=$(cat test.txt)
if [$(($n+8)) -ge 100]
then
echo 'terminated program' > test.txt
else
m=$(($n+3))
echo $m > test.txt
fi
You miss some spaces :
if [$(($n+8)) -ge 100]
->
if [ $(($n+8)) -ge 100 ]
But while using bash, prefer a modern solution, using bash arithmetic :
if (( n+8 >= 100 ))
or even
if ((n+8>=100))
Like #Gordon Davisson said in comments : arithmetic contexts like inside (( )) are one of the few places in bash where spaces aren't critical delimiters.

Verify amount in bash [duplicate]

This question already has answers here:
How can I compare two floating point numbers in Bash?
(22 answers)
Closed 5 years ago.
Currently i'm writing a script that checks whether an amount is higher than the other.
if [ "324,53" -gt "325,00" ]; then
echo "Amount is higher!"
else
echo "Amount is lower."
fi
Obviously i'm receiving the error: "integer expression expected" when running this. I tried to use sed 's/[^0-9]*//g'` but i need the full number again for further use. Problem is that I don't know the correct way how to do this. Can anybody give some advice?
Bash can't handle floating point numbers. You need to use an external tool, e.g. bc, but you need to use decimal point, not comma:
if (( $(bc <<< '324.53 > 325.00') )) ; then
bc prints 1 when the condition is true, and 0 when it's false. $(...) takes the output of the command, and (( ... )) interpret it as a number, returning false for zero and true for anything else.

Comparing strings composed from numbers

I'm trying to compare values of two variables both containing strings-as-numbers. For example:
var1="5.4.7.1"
var2="6.2.4.5"
var3="1-4"
var4="1-5"
var5="2.3-3"
var6="2.3.4"
Sadly, I don't even know where to start... Any help will be appreciated!
What I meant is how would I go about comparing the value of $var5 to $var6 and determine with one of them is higher.
EDIT: Better description of the problem.
You can use [[ ${str1} < ${str2} ]] style test. This should work:
function max()
{
[[ "$1" < "$2" ]] && echo $2 || echo $1
}
max=$(max ${var5} ${var6})
echo "max=${max}."
It depends of the required portability of the solution. If you don't care about that and you use a deb based distribution, you can use the dpkg --compare-versions feature.
However, if you need to run your script on distros without dpkg I would use following approach.
The value you need to compare consist of first (the first element) and the rest (all others). The first is usually called the head and the rest - tail, but I deliberately use names first and rest, to not confuse with head(1) and tail(1) tools available on Unix systems.
In case first($var1) is not equal to first($var2) you just compares those firsts elements. If firsts are equal, just recursively run the compare function on rest($var1) and rest($var2). As a border case you need to decide what to do if values are like:
var1 = "2.3.4"
var2 = "2.3"
and in some step you will compare empty and non-empty first.
Hint for implementing first and rest functions:
foo="2.3-4.5"
echo ${foo%%[^0-9][0-9]*}
echo ${foo#[0-9]*[^0-9]}
If those are unclear to you, read man bash section titled Parameter Expansion. Searching the manual for ## string will show you the exact section immediately.
Also, make sure, you are comparing elements numerically not in lexical order. For example compare the result of following commands:
[[ 9 > 10 ]]; echo $?
[[ 9 -gt 10 ]]; echo $?

BASH Arithmetic Issues

I'm working in BASH and I'm having an idiot moment right now. I've got a project I'm working on that I'm going to need to use some very basic arithmetic expressions and I just realized that a lot of my problems with it are because my variables are not updating. So I threw together a basic algorithm that increments a variable by another variable with a while loop until a certain number is reached.
counter=1
counter2=0
while [[ counter2 < 10 ]]; do
counter2=$(($counter2+$counter))
echo $counter
echo $counter2
done
I run the script. Does nothing. I set the < to > just for kicks and an infinite loop occurs with a repeated output of:
1
0
1
0
Forever and ever until I stop it. So it's obvious the variables are not changing. Why? I feel like such an idiot because it must be something stupid I'm overlooking. And why, when I have <, it also isn't an infinite loop? Why doesn't it print anything at all for that matter? If counter2 is always less than 10, why doesn't it just keep going on forever?
Thanks folks in advance.
EDIT: Well, I realize why it wasn't outputting anything for when the check is <... I should have been using $counter2 instead of just counter2 to get the actual value of counter2. But now it just outputs:
1
2
And that's it... I feel like such a derp.
If this is all bash (100% sure) then you could use declare -i in order to explicitly set type of your variables and then your code will be as simple as :
declare -i counter=1
declare -i counter2=0
while [[ $counter2 -lt 10 ]]; do
counter2=$counter2+$counter
echo $counter
echo $counter2
done
EDIT:
In bash, you can do arithmatic comparison with double paranethesis. So, your while can be written as:
while (($counter2 < 10)) ; do
Inside the $((...)), don't use the sigil ($).
counter2=$((counter2+counter))
In bash, you can use c-like for loops:
for (( counter2=0; counter2<10; counter2+=counter ))
do
echo $counter": "$counter2
done
Often you will find this construct more appealing to use:
for counter2 in {0..9}
do
echo $counter": "$counter2
done

Shell loops using non-integers?

I wrote a .sh file to compile and run a few programs for a homework assignment. I have a "for" loop in the script, but it won't work unless I use only integers:
#!/bin/bash
for (( i=10; i<=100000; i+=100))
do
./hw3_2_2 $i
done
The variable $i is an input for the program hw3_2_2, and I have non-integer values I'd like to use. How could I loop through running the code with a list of decimal numbers?
I find it surprising that in five years no one ever mentioned the utility created just for generating ranges, but, then again, it comes from BSD around 2005, and perhaps it wasn't even generally available on Linux at the time the question was made.
But here it is:
for i in $(seq 0 0.1 1)
Or, to print all numbers with the same width (by prepending or appending zeroes), use -w. That helps prevent numbers being sent as "integers", if that would cause issues.
The syntax is seq [first [incr]] last, with first defaulting to 1, and incr defaulting to either 1 or -1, depending on whether last is greater than or less than first. For other parameters, see seq(1).
you can use awk to generate your decimals eg steps of0.1
num=$(awk 'BEGIN{for(i=1;i<=10;i+=0.1)print i}')
for n in $num
do
./hw3_2_2 $n
done
or you can do it entirely in awk
awk 'BEGIN{cmd="hw3_2_2";for(i=1;i<=10;i+=0.1){c=cmd" "i;system(cmd) } }'
The easiest way is to just list them:
for a in 1.2 3.4 3.11 402.12 4.2 2342.40
do
./hw3_2_2 $a
done
If the list is huge, so you can't have it as a literal list, consider dumping it in a file and then using something like
for a in $(< my-numbers.txt)
do
./hw3_2_2 $a
done
The $(< my-numbers.txt) part is an efficient way (in Bash) to substitute the contents of the names file in that location of the script. Thanks to Dennis Williamson for pointing out that there is no need to use the external cat command for this.
Here's another way. You can use a here doc to include your data in the script:
read -r -d '' data <<EOF
1.1
2.12
3.14159
4
5.05
EOF
for i in "$data"
do
./hw3_2_2 "$i"
done
Similarly:
array=(
1.1
2.12
3.14159
4
5.05
)
for i in "${array[#]}"
do
./hw3_2_2 "$i"
done
I usually also use "seq" as per the second answer, but just to give an answer in terms of a precision-robust integer loop and then bc conversion to a float:
#!/bin/bash
for i in {2..10..2} ; do
x=`echo "scale=2 ; ${i}/10" | bc`
echo $x
done
gives:
.2
.4
.6
.8
1.0
bash doesn't do decimal numbers. Either use something like bc that can, or move to a more complete programming language. Beware of accuracy problems though.

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