Bash – Increment Decimal Variable with Leading Zero by +1

arithmeticbashnumeric datashell-script

I have a file in the name of Build.number with the content value 012 which I need to increment by +1.
So, I tried this

BN=$($cat Build.number)
BN=$(($BN+1))
echo $BN >Build.number

but here I am getting the value 11 when I am expecting 013.
Can anyone help me?

Best Answer

The leading 0 causes Bash to interpret the value as an octal value; 012 octal is 10 decimal, so you get 11.

To force the use of decimal, add 10# (as long as the number has no leading sign):

BN=10#$(cat Build.number)
echo $((++BN)) > Build.number

To print the number using at least three digits, use printf:

printf "%.3d\n" $((++BN)) > Build.number

Related Solutions

Bash – Increment a variable using a Bash script

An awk based solution could be:

awk '/^#define PF_BUILD_VERSION / {$3++} 1' infile >outfile  &&  mv outfile infile

Shell – String to integer in Shell

The regex below will extract the number of bytes, only the number:

contentlength=$(
  LC_ALL=C sed '
    /^[Cc]ontent-[Ll]ength:[[:blank:]]*0*\([0-9]\{1,\}\).*$/!d
    s//\1/; q' headers
)

After the above change, the contentlength variable will be only made of decimal digits (with leading 0s removes so the shell doesn't consider the number as octal), thus the 2 lines below will display the same result:

echo "$(($contentlength/9))"

echo "$((contentlength/9))"

Best Answer

Related Solutions

Bash – Increment a variable using a Bash script

Shell – String to integer in Shell

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