Describe the bug

You can't pass the pseudo-mode ! to maparg().

To Reproduce

Run this shell command:

vim -Nu NONE +'noremap! <c-a> <c-b>'

Run this Ex command:

:echo maparg('<c-a>', '!', 0, 1)

The output is an empty dictionary.

Expected behavior

The output is this dictionary:

{'lnum': 0, 'script': 0, 'simplified': 1, 'silent': 0, 'noremap': 1, 'lhs': '<C-A>', 'mode': '!', 'nowait': 0, 'expr': 0, 'sid': -3, 'rhs': '<c-b>', 'buffer': 0}

Environment

• Vim version: 8.2 Included patches: 1-812
• OS: Ubuntu 16.04.6 LTS
• Terminal: XTerm(322)

Additional context

I think it's an omission because when reading :h maparg(), one can see that {mode} can be any mode used in a mapping command:

{mode} can be one of these strings:
        "n"	Normal
        "v"	Visual (including Select)
        "o"	Operator-pending
        "i"	Insert
        "c"	Cmd-line
        "s"	Select
        "x"	Visual
        "l"	langmap |language-mapping|
        "t"	Terminal-Job
        ""	Normal, Visual and Operator-pending

• n = :nmap
• v = :vmap
• o = :omap
• i = :imap
• c = :cmap
• s = :smap
• x = :xmap
• l = :lmap
• t = :tmap
• '' = :map

For each mapping command, there is a corresponding flag which can be passed as a second argument to maparg(). Except for the :map! command; there is no corresponding ! flag which could be passed to maparg().

Because of this, you can't use mapset() to restore a mapping installed via :map!.

vim -Nu NONE -S <(cat <<'EOF'
    noremap! X Y
    let save_map = maparg('X', '!', 0, 1)
    unmap! X
    call mapset('!', 0, 1)
    call feedkeys('iX')
EOF
)
# result:   'X' is inserted
# expected: 'Y' is inserted

That's because the first {mode} argument of mapset() is the same as for maparg():

{mode} and {abbr} should be the same as for the call to |maparg()|.

So, since ! can't be used for maparg(), it can't be used for mapset() either.