Removing Leading 0 from a number causes value to change [duplicate] - bash

Question

This question already has answers here:
Shell equality operators (=, ==, -eq)
(4 answers)
Closed 1 year ago.
I do not understand what is happening here, based on Convert string into integer in bash script - "Leading Zero" number error, I am able to convert the string into integer. I can even do addition. However, when I compare the numbers it gives me the wrong comparison. Refer to example below
Here's an example
array=(02 08 10 11 23 52)
for i in 0 1 2 3 4 5
do
if [[ ${array[i]#0} > 15 ]]
then
echo "${array[i]#0} is greater than 15"
else
echo "${array[i]#0} is less than 15"
fi
done
Output
2 is greater than 15
8 is greater than 15
10 is less than 15
11 is less than 15
23 is greater than 15
52 is greater than 15

Answer 1

Either change [[ ]] to (( )) to use > or change the operator to -gt to do integer comparisons inside of [[ ]]
First:
#!/bin/bash
array=(02 08 10 11 23 52)
for i in 0 1 2 3 4 5
do
if (( ${array[i]#0} > 15 ))
then
echo "${array[i]#0} is greater than 15"
else
echo "${array[i]#0} is less than 15"
fi
done
Second:
#!/bin/bash
array=(02 08 10 11 23 52)
for i in 0 1 2 3 4 5
do
if [[ ${array[i]#0} -gt 15 ]]
then
echo "${array[i]#0} is greater than 15"
else
echo "${array[i]#0} is less than 15"
fi
done
If you use < > == inside of [ ] or [[ ]] is a string comparison operator. The string "02" is in fact less than the string "15":
% [[ 02 > 15 ]]; echo $?
1 # '1' means false and '0' means true
BUT you are stripping the leading "0" with #0:
% s="02"
% echo ${s#0}
2
So now you are comparing the STRING "2" to the STRING "15":
% [[ 2 > 15 ]]; echo $?
0
% [[ ${s#0} > 15 ]]; echo $?
0
Which give the surprising result you are seeing...
See this answer for an expanded set of examples.