Shell – How printf with Multiple Arguments Separated by Space Works

printfshell

I have the following printf function:

$ printf '%s %s %s\t%s\n' 100644 blob 8e1e f1.txt 100644 blob 9de7 f2.txt | git mktree

Can anyone please explain what it produces and why? I expected to have equal number of formatting options %s for each argument string but I have a lot more arguments strings here.

Best Answer

The format is reused as many times as needed to display all arguments. If there are too few arguments, the missing arguments are treated as empty strings.

Examples

Here is an example of a format that specifies two arguments but only one is provided:

$ printf '%s ; %s ;\n' a
a ;  ;

Here is the same format, this time provided with one too many arguments:

$ printf '%s ; %s ;\n' a b c
a ; b ;
c ;  ;

Here is the example from the question in which the format expects four arguments. Since eight arguments are provided the entire format is used twice:

$ printf '%s %s %s\t%s\n' 100644 blob 8e1e f1.txt 100644 blob 9de7 f2.txt
100644 blob 8e1e        f1.txt
100644 blob 9de7        f2.txt

Documentation

From man bash:

The format is reused as necessary to consume all of the arguments. If the format requires more arguments than are supplied, the extra format specifications behave as if a zero value or null string, as appropriate, had been supplied. The return value is zero on success, non-zero on failure.

Related Solutions

Bash – Printf formatting with variable format – what does this var reference

The code "%$width.${width}s\n" generates a format string that is suitable for consumption by printf

In the script that you posted, width has been assigned the value 55, so that both $width and ${width} are expanded by bash to 55: the entire first parameter to printf expands to %55.55s\n; this is the format %s, with a field width and precision specifiers that ask to print on exactly 55 characters. Given the value of variable divider at this point, this will simply print a line of 55 equal signs. A perhaps simpler way of printing the same thing would have been perl -e 'print "=" x 55, "\n"'.

The simplest form for the field width specifier is an integer: this asks printf to use at least this many characters for printing. If the corresponding parameter requires fewer characters than this to print, then the output is left-padded with spaces.

The simplest form for the precision specifier is a dot followed by an integer: when applied to %s, this sets the maximum number of characters to print. (It has a different meaning for numeric types.)

In response to a comment, I will also mention a little bit about shell variable expansion (the complete explanation can be found by searching for "parameter expansion" in the bash documentation, see also $VAR vs ${VAR} and to quote or not to quote):

When a variable, say x, has been set, then $x expands to the value of x. If that value contains whitespace, then the expansion will be several words. This is why it's important in the code above that, for example, "$format" is within double quotes: this forces the expansion to be a single word (otherwise, printf would see a first parameter %-15s, followed by an argument %8s, etc., instead of receiving the entire format string as a single parameter).
It is permissible to write ${x} instead of just $x to expand the variable x; in the case above, "${width}s", it is necessary to do so because if one had written "$widths", then bash would try to get the value of variable widths, which is unset, resulting in an empty expansion.

Bash – Handling arguments in specified order in /usr/bin/printf or Bash printf

You can make bash do it without invoking other processes by redefining printf as a function:

printf() {  # add support for numbered conversion specifiers
    local -a args
    local opt=
    case $1 in
    -v) opt=-v$2; shift 2;;
    -*) opt=$1; shift;;
    esac
    local format=$1; shift
    while [[ $format =~ ((^|.*[^%])%)([0-9]+)\$(.*) ]]
    do
        args=("${!BASH_REMATCH[3]}" "${args[@]}")
        format=${BASH_REMATCH[1]}${BASH_REMATCH[4]}
    done
    let ${#args[@]} && set -- "${args[@]}"
    builtin printf $opt "$format" "$@"
}

This works by replacing numbered conversion specifiers in the format string with unnumbered ones while using the numbers to select arguments for a modified argument list. If no numbered conversion specifiers are present, the printf builtin will see the function's arguments as-is. Should be a drop-in replacement with very little surprising behaviour.

The possibly surprising ordering in the statement that builds the new argument list, with new arguments pushed onto the front of the args[] array instead of being appended, is needed because bash regex matching is greedy; each time around the while loop, the numbered conversion specifier it finds will be the last one remaining in the format string.

Also note that the format string recycling that the printf builtin normally performs when passed more arguments than the format string contains conversion specifiers for doesn't happen when numbered conversion specifiers are present; arguments that don't correspond to a numbered specifier are simply ignored. Given that POSIX describes the mixing of numbered and unnumbered conversion specifiers as undefined, I think that's reasonable.