CODEWITHSUNDEEP
bashintegerprintfdecimaloctal
user2118915
user2118915

Reputation: 532

strange behavior when fixing integer lengths in bash

In my bash script, I set-up the following operations

year=0050
echo $(printf %04d $year)
>0040

I do not understand why 0040 is returned instead of 0050. I eventually found that to get the system to print 0050 correctly, I will have to do this instead.

year=50
echo $(printf %04d $year)
>0050

Are there any insights as to why the first case happens?

Upvotes: 1

Views: 54

Answers (1)

Benjamin W.
Benjamin W.

Reputation: 52112

It's because numbers with a leading zero are interpreted as octal by Bash, and octal 50 is decimal 40.

To fix it, you can either strip the zeros with a parameter expansion:

$ printf '%04d\n' "${year##+(0)}"
0050

I've dropped the echo $(...) construct and inserted a newline in the formatting string instead.

Notice that the +(0) pattern requires the extglob shell option (shopt -s extglob).

Alternatively (and more portably), you can convert the number with an arithmetic expansion first:

% printf '%04d\n' "$(( 10#$year ))"
0050

This uses the base#n notation to indicate that n (in our case: $year) is in base 10 and not octal.

Upvotes: 2

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