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

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.