\documentclass[12pt, russian]{article}
\tracingall
\tracingassigns=5
\tracingmacros=2
\tracingcommands=2
\tracingrestores=1
\tracingstats=0
\tracingoutput=10
\tracinggroups=10
\tracingonline0
\begin{document}
$$
\left\{
\hbox\bgroup
$
\vcenter\bgroup
\gdef\withlineskip{\hbox\bgroup$}
\gdef\endwithlineskip{$\egroup}
\gdef\\{%
\cr
\gdef\withlineskip{\hbox\bgroup \vbox\bgroup \hbox{}\hbox\bgroup$}%
\gdef\endwithlineskip{$\egroup\egroup\egroup}%
\gdef\\{\cr}%
}
\offinterlineskip
\halign{
\hfil\withlineskip # \endwithlineskip
&\withlineskip {}#{} \endwithlineskip
&\withlineskip # \endwithlineskip\hfil\ \vrule
&\withlineskip {} \cdot # \endwithlineskip\cr
g s_{12} + s_{21} & = & (x_{1}\psi_{1})(t) = \frac{1}{g}U_{1}(t) & (-v) \\
-s_{11} + v s_{12} + s_{22} & = & (x_{1}\psi_{2})(t) & g \\
g s_{22} - g s_{11} - v s_{21} & = & \frac{1}{g} (g U_{2}(t) - v U_{1}(t)) & (-1) \\
}
\egroup
$
\egroup
\right.
$$
\end{document}
我的意图在标题中。有人能解释一下 LaTeX 给出的错误吗？从.log文件中我了解到出了问题 -\\已被捕获到\halign模板条目中。无论如何，这总是会发生吗？如何解决我保存意图原理的问题？
对于那些寻求（几乎）正确输出的人，请编译这个。
\documentclass[12pt, russian]{article}
\begin{document}
$$
\left\{
\hbox\bgroup
$
\displaystyle
\vcenter\bgroup
\gdef\withlineskip{$\displaystyle}
\gdef\endwithlineskip{$}
\gdef\\{%
\cr%
}
\offinterlineskip
\halign{%
\hfil\withlineskip#\endwithlineskip
&\withlineskip{}#{}\endwithlineskip
&\withlineskip#\endwithlineskip\hfil\ \vrule
&\withlineskip{}\cdot#\endwithlineskip\hfil\cr
g s_{12} + s_{21} & = & (x_{1}\psi_{1})(t) = \frac{1}{g}U_{1}(t) &(-v)\\
-s_{11} + v s_{12} + s_{22} & = & (x_{1}\psi_{2})(t) &g\\
g s_{22} - g s_{11} - v s_{21} & = & \frac{1}{g} (g U_{2}(t) - v U_{1}(t)) &(-1)\\
}
\egroup
$
\egroup
\right.
$$
\end{document}
你注意到第三行和第二行发生冲突了吗？这就是我要解决的问题。（好吧，我的方法中又发现了一个错误，等有人向我解释错误后，我会改正的。）
更新
这就是我最初想要的：
\documentclass[12pt, russian]{article}
\begin{document}
\begin{equation}
\left\{
\hbox\bgroup
$
\vcenter\bgroup
% \openup1\jot
\edef\restoreinterlineskip{%
\baselineskip=\the\baselineskip
\lineskip=\the\lineskip
\lineskiplimit=\the\lineskiplimit
}
\gdef\withlineskip{%
\hbox\bgroup
\hbox\bgroup
\hbox\bgroup
$%
\displaystyle
}
\gdef\endwithlineskip{%
$%
\egroup
\egroup
\egroup
}
\gdef\nextrow{%
\cr
\noalign{\xdef\prevrowdepth{\the\prevdepth}\vskip -\prevrowdepth}%
\gdef\withlineskip{%
\hbox\bgroup
\restoreinterlineskip
\vbox\bgroup
\hbox{\vrule width 0pt height 0pt depth \prevrowdepth }%
\hbox\bgroup
$
\displaystyle
}%
\gdef\nextrow{%
\cr
\noalign{\xdef\prevrowdepth{\the\prevdepth}}%
}%
}
\offinterlineskip
\halign{
\hfil\withlineskip # \endwithlineskip
&\withlineskip {}#{} \endwithlineskip
&\withlineskip # \endwithlineskip\hfil\ \vrule
&\withlineskip {} \cdot # \endwithlineskip\cr
g s_{12} + s_{21} & = & (x_{1}\psi_{1})(t) = \frac{1}{g}U_{1}(t) & (-v) \nextrow
-s_{11} + v s_{12} + s_{22} & = & (x_{1}\psi_{2})(t) & g \nextrow
g s_{22} - g s_{11} - v s_{21} & = & \frac{1}{g} (g U_{2}(t) - v U_{1}(t)) & (-1) \nextrow
}
\egroup
$
\egroup
\right.
\Longrightarrow 0 = 0
\end{equation}
\end{document}
无需手动选择跨行跳过！:-)
答案1
问题在于 是\gdef之后第一列的一部分\cr。它们是在\hfil\whitlineskip扩展第一列的第一个之后处理的。因此\endwithlineskip第一列与 不匹配\withlineskip。
我不知道要求是什么。可能只是在行之间插入空白行。然后\noalign{\bigskip}解决\cr了它。但这是纯 TeX 方法，我不明白为什么它被 LaTeX 的命令包围。
答案2
这是您所追求的 (LaTeX 语法) 吗？
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\renewcommand{\arraystretch}{1.5}
\begin{array}{rl|l}
gs_{12} + s_{21} & = (x_1\psi_1)(t) = \dfrac1g U_1(t) & {}\cdot(-v)\\
-s_{11} + vs_{12} + s_{22} & = (x_1\psi_2)(t) & {}\cdot g\\
gs_{22} - gs_{11} - vs_{21} & = \dfrac1g (gU_2(t) - vU_1(t)) & {}\cdot (-1)\\
\end{array}
\]
\end{document}
使用纯格式的 TeX 原语，可以生成类似这样的内容：
$$
\left\{\,
\vcenter{
\let\DS\displaystyle
\mathsurround=0pt
\def\cr{\crcr\noalign{\nointerlineskip}}
\def\invisiblethingamabob{%
\vrule height\baselineskip depth.6\baselineskip width0pt}
\halign{\invisiblethingamabob $\DS\hfil #$& $\DS{}#\hfil\ \vrule $&
$\;\DS{}\cdot #\hfil$\crcr
gs_{12} + s_{21} &= (x_1\psi_1)(t) = {1\over g} U_1(t) & (-v)\cr
- s_{11} + vs_{12} + s_{22} &= (x_1\psi_2)(t) & g \cr
gs_{22} - gs_{11} - vs_{21} &= {1\over g} (gU_2(t) - vU_1(t)) & (-1)\cr
}}
\right.
$$
答案3
你为一些在 LaTeX 中相当简单的事情做了太多的工作。
\documentclass[12pt]{article}
\usepackage{array}
\begin{document}
\[
\left\{
\begin{array}{@{}r@{}>{{}}l |@{} >{{}\cdot}l@{}}
g s_{12} + s_{21} &= (x_{1}\psi_{1})(t) = \frac{1}{g}U_{1}(t) & (-v) \\[2ex]
-s_{11} + v s_{12} + s_{22} &= (x_{1}\psi_{2})(t) & g \\[2ex]
g s_{22} - g s_{11} - v s_{21} &= \frac{1}{g} (g U_{2}(t) - v U_{1}(t)) & (-1)
\end{array}
\right.
\]
\end{document}
不，即使你要求，我也不会删除这个答案。
你的宏为什么不起作用？你正在尝试以最复杂的方式做事。
这是一个在 Plain TeX 中或多或少直接的实现。
\def\frac#1#2{{\begingroup#1\endgroup\over#2}}
$$
\left\{
\vcenter{
\tabskip=0pt
\offinterlineskip
\halign{\strut\hfil$#$&${}#$\hfil&\hskip3pt\vrule#&${}\cdot#$\hfil\cr
g s_{12} + s_{21}& = (x_{1}\psi_{1})(t) = \frac{1}{g}U_{1}(t)&& (-v)\cr
\omit&& height1.5ex&\omit\cr
-s_{11} + v s_{12} + s_{22}& = (x_{1}\psi_{2})(t)&& g\cr
\omit&& height1.5ex&\omit\cr
g s_{22} - g s_{11} - v s_{21}&= \frac{1}{g} (g U_{2}(t) - v U_{1}(t))&& (-1)\cr
}
}
\right.
$$
\bye
