How can I include a variable in a Perl printf expression? - perl

Question

How can I include a variable in a printf expression?
Here's my example:
printf "%${cols}s", $_;
Where $cols is the number of columns
and $_ is a string.
The statement results in an "Invalid conversion" warning.
The problem ended up being that I forgot to chomp the variable. Gah. Thanks everyone.

Answer 1

Your interpolated variable $cols looks like its supposed to be a number, say 10, so
"%${cols}s"
should interpolate and be equivalent to
"%10s"
which is a valid format string.
If however $cols was something other than a number or valid format string, you'd get the warning.
For example, if:
$cols = "w";
that would result in "%ws" as a format string - giving the error you quote:
Invalid conversion in printf: "%w"
Valid format information can be found here.

Answer 2

I figured out your specific problem. Your code is correct. However, I suppose $cols might be a number read from user input, say like this:
my $cols = <STDIN>;
This works, and in numeric context $cols will appear to be a number, but the problem is that $cols isn't appearing in numeric context here. It's in string context, which means that instead of expanding to "%5s", your format string expands to "%5\ns". The newline there is mucking up the format string.
Change the code where you read $cols to this:
chomp(my $cols = <STDIN>);
See the documentation on chomp, as you may want to use it for other input reading as well.

Answer 3

Always use * in your format specifier to unambiguously indicate variable width! This is similar to the advice to use printf "%s", $str rather than printf $str.
From the perlfunc documentation on sprintf:
(minimum) width
Arguments are usually formatted to be only as wide as required to display the given value. You can override the width by putting a number here, or get the width from the next argument (with *) or from a specified argument (with e.g. *2$):
printf '<%s>', "a"; # prints "<a>"
printf '<%6s>', "a"; # prints "< a>"
printf '<%*s>', 6, "a"; # prints "< a>"
printf '<%*2$s>', "a", 6; # prints "< a>"
printf '<%2s>', "long"; # prints "<long>" (does not truncate)
If a field width obtained through * is negative, it has the same effect as the - flag: left-justification.
For example:
#! /usr/bin/perl
use warnings;
use strict;
my $cols = 10;
$_ = "foo!";
printf "%*s\n", $cols, $_;
print "0123456789\n";
Output:
foo!
0123456789
With the warnings pragma enabled, you'll see warnings for non-numeric width arguments.

Answer 4

Your current method should work
perl -e'my $cols=500; $_="foo"; printf "%${cols}s\n\n", $_;'

Answer 5

The following seems to work for me:
#!/bin/perl5.8 -w
use strict;
my $cols = 5;
my $a = "3";
printf "%${cols}d\n", $a;
yields
28$ ./test.pl
3
29$

Answer 6

I cannot reproduce your problem. The following code works fine:
use strict;
use warnings;
my $cols=40;
while (<>) {
printf "%${cols}s\n", $_;
}
It prints any input line using at least 40 columns of width.