I've been trying to figure out why this is happening with quotes and now I seem to be lost.

I'm setting a variable (in BASH) to hold a directory tree listing:
Listing var x gives:
But listing var x with it quoted, gives:
Why does quoting the variable generate the newline?

Thanks.

Your "echo" command gets a string from the var.
But it's default behavior is to suppress the interpretation of sequences like "\n".
So it does, what it is designed to: it concatenates all given strings (disregarding the contained newlines), inserts a space between the strings and outputs it.
This takes place with your "echo $x".

If you quote "$var" echo gets one big string containing newline characters and as it has nothing to concatenate it outputs the string, as it is and the contained "\n" are interpreted in a normal fashion.

Some remarks.
Never parse the output of "ls" "ps" and alike.
They depend on language settings and are not portable.
In fact, if using the output of "ls" you are NOT dealing with filenames, instead you are dealing with representations of filenames.
Use find instead.
"find . -type d " gives a newline separated list of all found directories.

And to avoid the quoting hell ALWAYS quote each and every var.
Use "$var" and never, never $var.

If the use of unquoted vars seems to fit, be sure: your script logic is wrong. Definitely!

Another point is, that there are loads of "echo"s out.
It could be a binary, a bash builtin or even an alias.
If you are going to examine this deeper take care to know, which kind of "echo" you are using.
"env echo" calls it from the path.
"echo" only god i.e. the sysadmin knows.
erm. Should know.

You are observing word splitting resulting from the parameter expansion of $x, itself.

From the bash man page:

So, word splitting usually happens to the output of a parameter expansion, except when it is in double quotes. Since newline is in $IFS, it gets stripped.

This isn't quite right.

Echo respects newlines in input strings:
The suppression of special chars (e.g. \n) has to do with when it is received literally to echo:
vs.

Maybe I did not write clear enough.

What you mentioned is right.
But this does not resolve the question asked.

Try this with the defined var x :

@uhelp You're right, I guess I did not actually answer the question. Also, od is a useful tool -- I had almost forgot about it. Glad you reintroduced me. :-)

So the answer here is that it doesn't generate the newline. Rather, by not quoting the variable, the shell performs word splitting on the output of the parameter expansion.

Thanks for all the responses. So a few weeks ago, I realized I did not know regex's as much as I thought I did. Now this. :-)

I'm using it in a shell script:

Of course, in the above, quoting $mydirs would cause the commandlist to fail.

I'll change to "find" once I'm clear on this.

Thanks again

I was going to say -- this is a bit clearer:

Sorry for the confusion, I meant once I was a bit clearer on the whole quoting stuff.

I was just asking if not quoting in the for..done loop was OK, since quoting the var definitely would not work.

Thank you. That's a more fool-proof method. I've gotten so accustomed to not using spaces in dir/file names anymore, I did not consider it.