我正在尝试创建一个使用嵌套序列的命令，例如 [[1,2,3],[4,5,6],...] 其中括号代表一个序列
我现在遇到的问题是无法将宏作为参数传递。希望得到帮助！
日志：
> ! Argument of \@sect has an extra }.
> <inserted text>
> \par
> l.42 \add{7}{8}{\textbf{9}}
>
> I've run across a `}' that doesn't seem to match anything.
> For example, `\def\a#1{...}' and `\a}' would produce
> this error. If you simply proceed now, the `\par' that
> I've just inserted will cause me to report a runaway
> argument that might be the root of the problem. But if
> your `}' was spurious, just type `2' and it will go away.
>
> Runaway argument?
> {\normalfont \Large \bfseries }{9}\ifx \reserved@a \@empty \let \check@icl \ETC
> .
> ! Paragraph ended before \@sect was complete.
> <to be read again>
> \par
> l.42 \add{7}{8}{\textbf{9}}
>
> I suspect you've forgotten a `}', causing me to apply this
> control sequence to too much text. How can we recover?
> My plan is to forget the whole thing and hope for the best.
>
> [1
梅威瑟：
\documentclass{article}
\usepackage{expl3}
\usepackage{tikz}
\usetikzlibrary{positioning}
\ExplSyntaxOn
\cs_generate_variant:Nn \seq_put_right:Nn {Ne}
\cs_generate_variant:Nn \seq_put_right:Nn {NV}
\seq_new:N \l__outer_seq
\seq_new:N \l__inner_seq
\int_new:N \l__last_node_int
\NewDocumentCommand\add{m m m}
{
\seq_clear:N \l__inner_seq
\seq_put_right:Ne \l__inner_seq {#1}
\seq_put_right:Ne \l__inner_seq {#2}
\seq_put_right:Ne \l__inner_seq {#3}
\seq_put_right:NV \l__outer_seq \l__inner_seq
}
\NewDocumentCommand\draw{o}
{
\int_set:Nn \l_tmpa_int {1}
\int_set:Nn \l_tmpb_int {\int_eval:n {\seq_count:N \l__outer_seq + 1}}
\int_zero:N \l__last_node_int
\begin{tikzpicture}[every~node/.style={
draw,
line~width=1pt,
minimum~height=30pt,
node~distance=-1pt,
transform~shape,
shape=rectangle,
}]
\int_do_while:nNnn {\l_tmpa_int} < {\l_tmpb_int}
{
\seq_pop:NN \l__outer_seq \l_tmpa_seq
\int_compare:nNnTF \l__last_node_int = 0
{
\int_incr:N {\l__last_node_int}
\node (a\int_use:N \l__last_node_int) {\seq_item:Nn \l_tmpa_seq {1}};
\node[right=of~{a\int_use:N \l__last_node_int}] (b\int_use:N \l__last_node_int) {\seq_item:Nn \l_tmpa_seq {2}};
\node[right=of~{b\int_use:N \l__last_node_int}] (c\int_use:N \l__last_node_int) {\seq_item:Nn \l_tmpa_seq {3}};
}
{
\int_incr:N {\l__last_node_int}
\node[below=of~{a\int_eval:n {\l__last_node_int - 1}}] (a\int_use:N \l__last_node_int) {\seq_item:Nn \l_tmpa_seq {1}};
\node[below=of~{b\int_eval:n {\l__last_node_int - 1}}] (b\int_use:N \l__last_node_int) {\seq_item:Nn \l_tmpa_seq {2}};
\node[below=of~{c\int_eval:n {\l__last_node_int - 1}}] (c\int_use:N \l__last_node_int) {\seq_item:Nn \l_tmpa_seq {3}};
}
\int_incr:N {\l_tmpa_int}
}
\end{tikzpicture}
\tl_clear:N \l__last_node_tl
}
\ExplSyntaxOff
\begin{document}
\section{MWE}
\add{1}{2}{3}
\add{4}{5}{6}
% Error
% \add{7}{8}{\textbf{9}}
\draw
\end{document}
答案1
您不能将序列存储在序列中：向您提出的代码仅存储一个变量，该变量将始终指向其当前值。
但是，您可以使用合适的分隔符存储序列项，然后在需要时重新构建它。
\int_step_inline:nn我使用比 while 循环更简单的代码来简化代码。我还根据推荐的指导方针重命名了变量。
另一个问题是使用\seq_put_right:Neon \textbf{9}，这是错误的。我将其替换为\seq_put_right:Nn。
\documentclass{article}
\usepackage{expl3}
\usepackage{tikz}
\usetikzlibrary{positioning}
\ExplSyntaxOn
\quark_new:N \q_nyannyan_sep
\cs_generate_variant:Nn \seq_put_right:Nn {Ne}
\seq_new:N \l__nyannyan_outer_seq
\seq_new:N \l__nyannyan_inner_seq
\int_new:N \l__nyannyan_last_node_int
\NewDocumentCommand\add{m m m}
{
\seq_clear:N \l__nyannyan_inner_seq
\seq_put_right:Nn \l__nyannyan_inner_seq {#1}
\seq_put_right:Nn \l__nyannyan_inner_seq {#2}
\seq_put_right:Nn \l__nyannyan_inner_seq {#3}
\seq_put_right:Ne \l__nyannyan_outer_seq
{
\seq_use:Nn \l__nyannyan_inner_seq { \q_nyannyan_sep }
}
}
\NewDocumentCommand\drawpicture {}
{
\int_zero:N \l__nyannyan_last_node_int
\begin{tikzpicture}[
every~node/.style={
draw,
line~width=1pt,
minimum~height=30pt,
node~distance=-1pt,
transform~shape,
shape=rectangle,
}
]
\int_step_inline:nn { \seq_count:N \l__nyannyan_outer_seq }
{
% reconstitute the stored sequences
\seq_pop_left:NN \l__nyannyan_outer_seq \l_tmpa_tl
\seq_set_split:NnV \l__nyannyan_inner_seq { \q_nyannyan_sep } \l_tmpa_tl
\int_compare:nNnTF { \l__nyannyan_last_node_int } = { 0 }
{
\int_incr:N \l__nyannyan_last_node_int
\node (a\int_use:N \l__nyannyan_last_node_int) {\seq_item:Nn \l__nyannyan_inner_seq {1}};
\node[right=of~{a\int_use:N \l__nyannyan_last_node_int}]
(b\int_use:N \l__nyannyan_last_node_int) {\seq_item:Nn \l__nyannyan_inner_seq {2}};
\node[right=of~{b\int_use:N \l__nyannyan_last_node_int}]
(c\int_use:N \l__nyannyan_last_node_int) {\seq_item:Nn \l__nyannyan_inner_seq {3}};
}
{
\int_incr:N \l__nyannyan_last_node_int
\node[below=of~{a\int_eval:n {\l__nyannyan_last_node_int - 1}}]
(a\int_use:N \l__nyannyan_last_node_int) {\seq_item:Nn \l__nyannyan_inner_seq {1}};
\node[below=of~{b\int_eval:n {\l__nyannyan_last_node_int - 1}}]
(b\int_use:N \l__nyannyan_last_node_int) {\seq_item:Nn \l__nyannyan_inner_seq {2}};
\node[below=of~{c\int_eval:n {\l__nyannyan_last_node_int - 1}}]
(c\int_use:N \l__nyannyan_last_node_int) {\seq_item:Nn \l__nyannyan_inner_seq {3}};
}
}
\end{tikzpicture}
}
\ExplSyntaxOff
\begin{document}
\section{MWE}
\add{1}{2}{3}
\add{4}{5}{6}
\add{7}{8}{\textbf{9}}
\drawpicture
\end{document}
