The printf command

incomplete Stranger, this is a very big topic that needs experience - please extend the descriptions and correct the details if you can!

Attention: This is about the Bash-builtin command printf - however, the description should be nearly identical for an external command that follows POSIX®.

Unlike other documentations, I don't want to redirect you to the manual page for the printf() C function family. However, if you're more experienced, that should be the most detailed description for the format strings and modifiers.

Due to mutual exclusive historical implementations of the echo command, POSIX® recommends to use printf rather than echo.

The printf command provides a method to print preformatted text similar to the printf() system interface (C function). It's meant as successor for echo and has far more features and possibilities.

Beside other reasons, POSIX® has a very good argument to recommend it: Both historical main flavours of the echo command are mutual exclusive, they collide. A “new” command had to be invented to solve the issue.

printf <FORMAT> <ARGUMENTS...>

The text format is given in <FORMAT>, while all arguments the formatstring may point to are given after that, here, indicated by <ARGUMENTS...>.

Thus, a typical printf-call looks like:

printf "Surname: %s\nName: %s\n" "$SURNAME" "$LASTNAME"

The -v Option can't assign directly to array indexes in Bash versions older than Bash 4.1

Of course in shell-meaning the arguments are just strings, however, the common C-notations plus some additions for number-constants are recognized to give a number-argument to printf:

If more arguments than format specifiers are present, then the format string is re-used until the last argument is interpreted. If less format specifiers than arguments are present, then number-formats are set to zero, while string-formats are set to null (empty).

Also, to minimize surprises, when printf expects an argument, give it one, not more. I'm talking about shell word splitting, please read this article if you don't know what I mean.

printf "%d\n" 08

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The format string interpretion is derived from the C printf() function family. Only format specifiers that end in one of the letters diouxXfeEgGcs are recognized.

To print a literal % (percent-sign), use %% in the format string.

Again: Every format specifier expects an associated argument provided!

Some of the mentioned format specifiers can modify their behaviour by getting a format modifier:

To be more flexible in the output of numbers and strings, the printf command allows format modifiers. These are specified between the introducting % and the character that specifies the format:

printf "%50s\n" "This field is 50 characters wide..."

The “alternative format” modifier #:

The precision for a floating- or double-number can be specified by using .<DIGITS>, where <DIGITS> is the number of digits for precision. If <DIGITS> is an asterisk (*), the precision is read from the argument that precedes the number to print, like (prints 4,3000000000):

printf "%.*f\n" 10 4,3

For strings, the precision specifies the maximum number of characters to print (i.e. the maximum field width). For integers, it specifies the number of digits to print (zero-padding!).

marks a code that may not be present, because it's not standardized by POSIX®.

Code	Description
\\	Prints the character \ (backslash)
\a	Prints the alert character (ASCII code 7 decimal)
\b	Prints a backspace
\f	Prints a form-feed
\n	Prints a newline
\r	Prints a carriage-return
\t	Prints a horizontal tabulator
\v	Prints a vertical tabulator
\<NNN>	Interprets <NNN> as octal number and prints the corresponding character from the character set
\0<NNN>	same as \<NNN>
\x<NNN>	Interprets <NNN> as hexadecimal number and prints the corresponding character from the character set (3 digits)
\u<NNNN>	same as \x<NNN>, but 4 digits
\U<NNNNNNNN>	same as \x<NNN>, but 8 digits

• print the decimal representation of a hexadecimal number (preserve the sign)
  • printf ”%d\n” 0×41
  • printf ”%d\n” -0×41
  • printf ”%+d\n” 0×41
• print the octal representation of a decimal number
  • printf ”%o\n” 65
  • printf ”%05o\n” 65 (5 characters width, padded with zeros)
• this prints a 0, since no argument is specified
  • printf ”%d\n”
• print the code number of the character A
  • printf ”%d\n” \'A
  • printf ”%d\n” ”'A”
• Generate a greeting banner and assign it to the variable GREETER
  • printf -v GREETER “Hello %s” “$LOGNAME”
• Print a text at the end of the line, using tput to get the current line width
  • printf ”%*s\n” $(tput cols) “Hello world!”

This small loop prints all numbers from 0 to 127 in

• decimal
• octal
• hex

for ((x=0; x <= 127; x++)); do
  printf '%3d | %04o | 0x%02x\n' "$x" "$x" "$x"
done

This code here will take a common MAC address and rewrite it into a well-known format (regarding leading zeros or upper/lowercase of the hex digits, ...):

the_mac="0:13:ce:7:7a:ad"

# lowercase hex digits
the_mac="$(printf "%02x:%02x:%02x:%02x:%02x:%02x" 0x${the_mac//:/ 0x})"

# or the uppercase-digits variant
the_mac="$(printf "%02X:%02X:%02X:%02X:%02X:%02X" 0x${the_mac//:/ 0x})"

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This code was found in Solaris manpage for echo(1).

Solaris version of /usr/bin/echo is equivalent to:

printf "%b\n" "$*"

Solaris /usr/ucb/echo is equivalent to:

if [ "X$1" = "X-n" ]
then
     shift
     printf "%s" "$*"
else
     printf "%s\n" "$*"
fi

Working off the replacement echo, here is a terse implementation of prargs:

printf '"%b"\n' "$0" "$@" | nl -v0 -s": "

A small trick: Combining printf and parameter expansion to draw a line

length=40
printf -v line '%*s' "$length"
echo ${line// /-}

length=40
eval printf -v line '%.0s-' {1..$length}

• External: Greg's BashFAQ 18: How can I use numbers with leading zeros in a loop, e.g. 01, 02?