Is it possible to add an "or" condition to a \def?

Question

Is it possible to have something like

\def\fiveorsix{5,6 OR 6,5}

So that the following works:

\documentclass{standalone}

\def\fiveorsix{5,6 OR 6,5}

\makeatletter
\newcommand{\test}[1][]{
    \def\test@a{#1}
    \ifx\test@a\fiveorsix
        the optional argument is 5,6 or 6,5
    \else
        the optional argument is not 5,6 or 6,5
    \fi}
\makeatother

\begin{document}
\test[5,6]
\end{document}

(obviously this does not work)

Or is there a better approach to achieve my goal?

edit: I had to update the MnWE because the first version was actually too minimal.

Answer 1

\documentclass{article}
\usepackage{expl3}[2013/07/24]
\ExplSyntaxOn
\newcommand{\test}[1][]{
  \str_case:nnTF {#1}
    {
      { 5,6 } { } % within those braces you could put code specific to the 5,6 case
      { 6,5 } { } % within those braces you could put code specific to the 6,5 case
    }
    { the~argument~is~5,6~or~6,5 }
    { the~argument~is~neither~5,6~nor~6,5 }
}
\ExplSyntaxOff
\begin{document}
\test[123]
\test[5,6]
\test[6,5]
\end{document}

Given the comments on your question, it seems that the optional argument is expected to be a pair of items separated by a comma. You may want to consider using xparse's \SplitArgument, which can split an argument at a given delimiter, with error reporting if the number of delimiters is not as expected. It also removes spaces around each item.

Answer 2

You could use the etoolbox (needs e-TeX):

\documentclass{article}

\usepackage{etoolbox}

\newcommand{\test}[1][]{%
    \ifboolexpr{
        test{\ifstrequal{#1}{5,6}} or
        test{\ifstrequal{#1}{6,5}
        }
    }
    {the optional argument is 5,6 or 6,5}
    {the optional argument is not 5,6 or 6,5}}

\begin{document} 
\test[3,2]

\test[5,6]
\end{document}

Results in:

the optional argument is not 5,6 or 6,5
the optional argument is 5,6 or 6,5

You may use other tests like \ifnumcomp. For details look at the etoolbox manual.

Answer 3

Not exactly as required, but the idea should be clear:

\documentclass{article}

\begin{document}

\def\fivesix#1{\ifcase#1 the  argument is not 5 or 6
\or the  argument is not 5 or 6
\or the  argument is not 5 or 6
\or the  argument is not 5 or 6
\or the  argument is not 5 or 6
\or the  argument IS 5 or 6
\or the  argument IS 5 or 6
\else  the  argument is not 5 or 6\fi}

\fivesix{1}

\fivesix{5}

\end{document}

Answer 4

Considering that you asked about the definition rather then the comparison, the exact answer to your question should be that it is not possible (in TeX). The macro expansion has to be unique. To achive this you have to specify in the definition under which circumstances the \fiveorsix macro should expand to '5,6' or '6,5'.

As far as you asked about alternate approaches to test whether the macro argument expands to '5,6' or '6,5' - and alredy got a bunch of answers - I add this generic solution for completeness:

\documentclass{article}

\def\fivsix{5,6}
\def\sixfiv{6,5}

\makeatletter
\newcommand{\test}[1][]{%
    \def\test@a{#1}
    \ifnum
      \ifx\test@a\fivsix 1\else\ifx\test@a\sixfiv 1\else 0\fi\fi
      =1
        the optional argument is 5,6 or 6,5
    \else
        the optional argument is not 5,6 and not 6,5
    \fi}
\makeatother

\begin{document}
\test[4,5]\par
\test[5,6]\par
\test[6,5]
\end{document}