General ⇒ expl3: don't understand behavior of tl_if_novalue:nTF

[erwann]

Before opening a ticket bug I'd like to make sure it is one indeed.

Code: Select all

Code, edit and compile here:

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\documentclass{minimal}
\usepackage{expl3}
\ExplSyntaxOn
\begin{document}
\tl_if_novalue:nTF{\c_novalue_tl}{PASS}{FAIL} % prints FAIL
\ExplSyntaxOff
\end{document}
 
 
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Run LaTeX here

[zauguin]

Actually this is the expected behaviour even if the documentation is a bit misleading:

\tl_if_novalue:nTF tests if the token list is exactly equal to \c_novalue_tl, meaning equal to the special token list which is stored inside \c_novalue_tl. This is not the same as {c_novalue_tl}, the token list consisting of exactly one token, \c_novalue_tl itself.

So if you would use a V style argument instead to compare the value of \c_novalue_tl, you get the expected behaviour:

Code: Select all

Code, edit and compile here:

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\documentclass{minimal}
\usepackage{expl3}
\ExplSyntaxOn
\begin{document}
\cs_generate_variant:Nn \tl_if_novalue:nTF { VTF }
\tl_if_novalue:VTF \c_novalue_tl {PASS}{FAIL} % prints PASS
\ExplSyntaxOff
\end{document}
 
 
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Run LaTeX here

In most actual use cases the normal behaviour is fine: The novalue marker is passed mostly passed as an argument to your macro, then you can write e.g. \tl_if_novalue:nTF{#1}{PASS}{FAIL} because the argument is not wrapped in a macro/variable.