如何在环境代码框内开始新页面？

Question

如果我把所有代码都写在一个框里codebox，页面就会满满的。如果我把代码写在两个框里，行数计数器就会从 1 重新开始。

我怎样才能保存行计数器？

Answer 1

这是一个初步版本，至少用于手动换行，使用\codeboxcontinuedtrue和\codeboxcontinuedfalse

最后一行号始终存储在 a 的末尾，如果计算结果为 ，则codebox可能会重新注入到计数器中。codelinenumber\ifcodeboxcontinuedtrue

\documentclass{article}

\usepackage{clrscode3e}
\usepackage{xpatch}

\newcounter{lastcodelinenumber}

\newif\ifcodeboxcontinued


\xapptocmd{\endcodebox}{%
  \setcounter{lastcodelinenumber}{\value{codelinenumber}}%
  \global\codeboxcontinuedfalse
}{}{}
\xapptocmd{\codebox}{%
  \ifcodeboxcontinued
  \setcounter{codelinenumber}{\value{lastcodelinenumber}}%
  \fi
}{}{}
\newcommand\continuecodebox{\global\codeboxcontinuedtrue}

\begin{document}

\begin{codebox}
\Procname{$\proc{Insertion-Sort}(A)$}
\li \For $j \gets 2$ \To $\attrib{A}{length}$
\li     \Do
$\id{key} \gets A[j]$
\li         \Comment Insert $A[j]$ into the sorted sequence
$A[1 \twodots j-1]$.
\li         $i \gets j-1$
\li         \While $i > 0$ and $A[i] > \id{key}$
\li             \Do
$A[i+1] \gets A[i]$
\li                 $i \gets i-1$
\End
\li         $A[i+1] \gets \id{key}$
\End
\end{codebox}
\clearpage % a manual break that would be necessary

\continuecodebox
\begin{codebox}
\li \For $j \gets 2$ \To $\attrib{A}{length}$
\li     \Do
$\id{key} \gets A[j]$
\li         \Comment Insert $A[j]$ into the sorted sequence
$A[1 \twodots j-1]$.
\li         $i \gets j-1$
\li         \While $i > 0$ and $A[i] > \id{key}$
\li             \Do
$A[i+1] \gets A[i]$
\li                 $i \gets i-1$
\End
\li         $A[i+1] \gets \id{key}$
\End
\li \For $j \gets 2$ \To $\attrib{A}{length}$
\li     \Do
$\id{key} \gets A[j]$
\li         \Comment Insert $A[j]$ into the sorted sequence
$A[1 \twodots j-1]$.
\li         $i \gets j-1$
\li         \While $i > 0$ and $A[i] > \id{key}$
\li             \Do
$A[i+1] \gets A[i]$
\li                 $i \gets i-1$
\End
\li         $A[i+1] \gets \id{key}$
\End
\li \For $j \gets 2$ \To $\attrib{A}{length}$
\li     \Do
$\id{key} \gets A[j]$
\li         \Comment Insert $A[j]$ into the sorted sequence
$A[1 \twodots j-1]$.
\li         $i \gets j-1$
\li         \While $i > 0$ and $A[i] > \id{key}$
\li             \Do
$A[i+1] \gets A[i]$
\li                 $i \gets i-1$
\End
\li         $A[i+1] \gets \id{key}$
\End
\end{codebox}
\end{document}