xargs

Quotes treated differently

xargs treats quotes differently than the shell (bash).

$ printf '"foo"bar"' | xargs printf %s\\n
xargs: unmatched double quote; by default quotes are special
to xargs unless you use the -0 option

Excerpt from man 1 xargs

[source,text

-0
--null

  Input items are terminated by a null character instead of by
  whitespace, and the quotes and backslash are not special (every
  character is taken literally). Disables the end-of-file string,
  which is treated like any other argument. Useful when input items
  might contain white space, quote marks, or backslashes. The GNU find
  (and from POSIX Issue 8, IEEE Std 1003.1, 2024) -print0 option
  produces input suitable for this mode.

Therefore:

$ printf '"foo"bar"' | xargs --null printf %s\\n
"foo"bar"

And, if we escape a quote with a backslash, nothing changes from the output above:

$ printf '"foo\"bar"' | xargs --null printf %s\\n
"foo"bar"

Note that the backslash is not printed. As far as I understand it, when a backslash scape escapes a character that is not treated in a special way, the escape does nothing.