How do I specify the position of a width parameter?

user_2678429 0
Viewed 127

Apologies if this has been posted before. I've been searching all over and can't find an answer.

According to man printf, "FORMAT controls the output as in C printf." It refers to printf(3).

According to man 3 printf, you can specify a variable width in a given position. It says:

Instead of a decimal digit string one may write … "*m$" (for some decimal integer m) to specify that the precision is given in the m-th argument … which must be of type int.

This is the part where I'm struggling. To give a simple example, suppose I wish to print a string with width 14.

$ printf '[%14s]\n' Something
[     Something]

I can use a variable instead:

$ WIDTH=14
$ printf '[%*s]\n' ${WIDTH} Something
[     Something]

The tricky part comes when I want to tell printf that the width argument is in a different position. To keep things simple for this example, I'll leave it where it is, in position 1. Following the instructions, I write the following.

Built-in version:

$ printf '[%*1$s]\n' ${WIDTH} Something
bash: printf: `1': invalid format character

Version in /usr/bin:

$ /usr/bin/printf '[%*1$s]\n' ${WIDTH} Something
[/usr/bin/printf: %*1: invalid conversion specification

Even using the example from the manual gives an error.

Built-in version:

$ printf '%2$*1$d' 6 34
bash: printf: `$': invalid format character

Version in /usr/bin:

$ /usr/bin/printf '%2$*1$d' 6 34
/usr/bin/printf: %2$: invalid conversion specification

As you can see, I get an error every time. I have struggled to see what I'm doing wrong, and I simply cannot find any example online.

How should this be formatted, please, or is the manual just wrong?

  • Lubuntu 18.04
  • GNU bash, version 4.4.20(1)-release (x86_64-pc-linux-gnu)
  • builtin printf: unknown version
  • /usr/bin/printf printf (GNU coreutils) 8.28

Thank you

printf linux bash
2 Answers

No, the extension that comes from Single UNIX Specification to C printf function to specify the position of positional parameter with $ is not supported in bash printf and I doubt it's supported in any of shells printf implementations. I believe the reason is that it requires additional work from the developers, while it's very rarely used extension.

is the manual just wrong?

FORMAT controls the output as in C printf.

You are reading the documentation for GNU version of printf, taken from the amazing Linux man-pages project. I would say C printf is documented in the C standard.

user_90727530

If your shell's printf doesn't support parameter indexing, you can still call Perl to the rescue!

$ perl -we 'printf q([%1$*2$d]), @ARGV' 36 6
[    36]

See printf and, more importantly, sprintf.

user_10306750
Related