Why does Vim on windows use \n for searching?

Question

So I was changing code from

foo()
{

to

foo() {

and I noticed that the searching pattern required me to searc

Answer 1

The syntax for the search part and replacement part are arbitrarily different. Some of the same codes are re-used to mean different things. Yeah, it's confusing.

    | How to type         | In search, means:       | In replacement, means:
----------------------------------------------------------------------------
\n  | \n                  | End-of-line             |  0x0
^@  | CTRL-V CTRL-J       |  0x0               |  0x0
\r  | \r                  | Carriage return 0xD     | "Break the line here"
^M  | CTRL-Enter          | Carriage return 0xD     | "Break the line here"
\^M | \ CTRL-V CTRL-ENTER | \ + carriage return 0xD | Carriage return 0xD

When searching, depending on your platform, the 0xD may be hidden, considered part of "newline", so yeah... you can always return sanity to your files and force all of the carriage returns displayed by opening a file and doing:

:e ++ff=unix

Similarly when replacing, "break the line here" does different things depending on your platform. It might insert 0xA or 0xD 0xA etc.

In case this isn't all already bad enough:

Technical detail:               *NL-used-for-Nul*
 characters in the file are stored as  in memory.  In the display
they are shown as "^@".  The translation is done when reading and writing
files.  To match a  with a search pattern you can just enter CTRL-@ or
"CTRL-V 000".  This is probably just what you expect.  Internally the
character is replaced with a  in the search pattern.  What is unusual is
that typing CTRL-V CTRL-J also inserts a , thus also searches for a 
in the file.  {Vi cannot handle  characters in the file at all}

                        *CR-used-for-NL*
When 'fileformat' is "mac",  characters in the file are stored as 
characters internally.  In the text they are shown as "^J".  Otherwise this
works similar to the usage of  for a .

When working with expression evaluation, a  character in the pattern
matches a  in the string.  The use of "\n" (backslash n) to match a 
doesn't work there, it only works to match text in the buffer.

Unless dealing with literal 0x0 or 0xD characters, it's agood a rule of thumb to always stick with \n for search and \r for replacement, as you've probably figured out.

See also:

:h /\n
:h /\r
:h s/\n
:h s/\r
:h s/\