cut -- line with no delimiters

I just discovered, to my dismay, the following part of the cut man page:

-f, --fields=LIST
select only these fields;
also print any line that contains no delimiter character, unless the -s option is specified

The -s option toggles the printing of lines with no delimiters.

In most cases, I would like lines with no delimiters to be considered as lines with a single field, so that:

$ echo "a" | cut -d' ' -f 1

would output a,
and

$ echo "a" | cut -d' ' -f 2

would output an empty line.

Is there an alternate version of cut where this is possible, or an option I don't know about?
As far as I understand, there is no option of cut that would return an empty line in the second case: if I use the -s option there is no line of output.

I can always use awk but in many cases this is not practical for me.

using awk, you can better handle these kind of situations.

Thanks, but as I said in many cases awk is not as practical as cut, so I was hoping for a solution using cut or something very similar.

For instance, I don't think there's an easy way in awk of printing all files, starting from the second for instance? (the equivalent of -f2- in cut)

i guess cut is not your way. i would use awk in cases like these.