How to print value of register to the standard output?

Question

To print the whole buffer, you can use %p, how do I print value of register to the standard output?

Here is example how I would print the whole buffer:

$ ex +'norm dw' +'%p' -scq! <(echo foo bar)
bar

So I'm looking for %p equivalent printing way for registers, for example to print the last deletion ("").

For example I've many columns in the file, and I want to delete selected column using ex and print what was deleted. — kenorb, Oct 16, 2015 at 15:57
E.g. This would remove 3rd column: ex +'%norm 2WdW' '+%p' -scq! <(echo 1 2 3 4 5) in shell, but I want opposite, to print what was actually deleted. — kenorb, Oct 16, 2015 at 16:00
@romainl The awk is more stream oriented, so there are some limitations in more complex parsing scenarios. On the other side awk is quicker, as I did some benchmarking recently, but I want to learn ex/vi more efficiently. — kenorb, Oct 16, 2015 at 22:04

kenorb · Accepted Answer · 2015-10-25 20:01:48Z

Echo command can print the value of any register:

:echon @"

This output can be redirected to the file (or standard output in this case) by using

:redir > file
:echon @"
:redir END

So the working example is as below:

$ ex +'norm dw' +'redir>>/dev/stdout|echon @-' -scq! <(echo foo bar)
foo

Where @- register stores the last text that you deleted or changed.

^{Note: The /dev/stdout could not exist on some non-Unix systems.}

Or workaround would be to replace the entire buffer with register and print it, like:

ex +'norm dw' +'%d|put-|%p' -scq! <(echo foo bar)

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