Stupid error caused by doing stupid things

A quick attempt with xparse:

\documentclass{article}
\usepackage{xparse,mathrsfs}
\NewDocumentCommand\optional{mO{}mm}
 {%
  \NewDocumentCommand{#1}{o#2}
   {\IfNoValueTF{##1}{#4}{#3}}%
 }
\NewDocumentCommand\starredcommand{O{s}mmm}
 {%
  \NewDocumentCommand{#2}{#1}
   {\IfBooleanTF{##1}{#4}{#3}}%
 }

\optional{\foo}{F(#1)}{F}
\optional{\baz}[m]{Baz with opt #1 and #2}{Baz without opt and #2}
\starredcommand{\barr}{\mathcal{B}}{\mathscr{B}}
\starredcommand[t{+}]{\carr}{\mathcal{C}}{\mathscr{C}}

\begin{document}
$\foo$

$\foo[x]$

$\foo[\barr*]$

$\foo[\barr]$

$\foo[\carr+]$

$\foo[\carr]$

\baz{x}

\baz[Y]{x}
\end{document}

However you have to change the way arguments are specified. You see in the example how to use a + instead of the *; for \optional and how to define \baz with a mandatory argument (the optional one is always present).

Here is how you can have your \starredcommand with arguments

\documentclass{article}
\usepackage{amsmath}
\usepackage{mathrsfs}
\makeatletter
% Following 3 lines thanks to Prof. Enrico Gregorio, from:
% http://tex.stackexchange.com/questions/42318/
%   removing-a-backslash-from-a-character-sequence
\begingroup\lccode`\|=`\\
\lowercase{\endgroup\def\removebs#1{\if#1|\else#1\fi}}
\newcommand{\@macro@name}[1]{\expandafter\removebs\string#1}
%
\newcommand\starredcommand[5]{%
  \gdef\mname{\@macro@name{#1}}%
  \newcommand#1{\@ifstar{\csname @\mname star\endcsname}
                        {\csname @\mname\endcsname}}
  \expandafter\newcommand\csname @\mname\endcsname[#2]{#3}
  \expandafter\newcommand\csname @\mname star\endcsname[#4]{#5}
}
\makeatother
\begin{document}

\starredcommand{\barr}{1}{\mathcal #1}{1}{\mathscr #1}

\begin{align}
    \barr{C} &\neq \barr*{F}
\end{align}

\end{document}

FOLLOW UP: Here is a version of your \optional command:

\usepackage{ifthen}
% ...
\def\optional#1[#2]#3#4{%
  \newcommand#1[#2][]{\ifthenelse{\equal{##1}{}}{#4}{#3}}%
}

though it is not ideal. It does allow, per the user's desire, for \barr to be an argument of foo, except that the \barr must be protected, as in

\optional{\foo}[1]{F(#1)}{F}
\( \foo = \foo[\protect\barr{Q}] \neq \foo[\protect\barr*{R}]\)