如何定義將巨集的定義作為參數的宏

Question 1

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\elliptic}{O{i}mmmm}
 {% #1 = item to substitute
  % #2 = main term
  % #3 = first index
  % #4 = last index
  % #5 = operation
  \group_begin:
  \lucas_elliptic:nnnnn { #1 } { #2 } { #3 } { #4 } { #5 }
  \group_end:
 }
\tl_new:N \l__lucas_elliptic_term_tl
\cs_generate_variant:Nn \cs_set:Nn { NV }
\cs_new:Nn \lucas_elliptic:nnnnn
 {
  \tl_set:Nn \l__lucas_elliptic_term_tl { #2 }
  \regex_replace_all:nnN
   { #1 } % search
   { \cB\{\cP\#1\cE\} } % replace
   \l__lucas_elliptic_term_tl % what to act on
  \cs_set:NV \__lucas_elliptic_term:n \l__lucas_elliptic_term_tl
  \__lucas_elliptic_term:n { #3 }
  #5 \dots #5
  \__lucas_elliptic_term:n { #4 }
 }
\ExplSyntaxOff

\begin{document}

$\elliptic{x^{i}}{0}{5}{+}$

$\elliptic{x_{i}}{0}{5}{+}$

$\elliptic[k]{(x_{k}+y_{k}i)}{1}{n}{}$

\end{document}

如果您願意使用#1佔位符，則可以簡化：

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\elliptic}{mmmm}
 {% #1 = main term
  % #2 = first index
  % #3 = last index
  % #4 = operation
  \group_begin:
  \lucas_elliptic:nnnn { #1 } { #2 } { #3 } { #4 }
  \group_end:
 }
\cs_new:Nn \lucas_elliptic:nnnn
 {
  \cs_set:Nn \__lucas_elliptic_term:n { #1 }
  \__lucas_elliptic_term:n { #2 }
  #4 \dots #4
  \__lucas_elliptic_term:n { #3 }
 }
\ExplSyntaxOff

\begin{document}

$\elliptic{x^{#1}}{0}{5}{+}$

$\elliptic{x_{#1}}{0}{5}{+}$

$\elliptic{(x_{#1}+y_{#1}i)}{1}{n}{}$

\end{document}

Answer

這是我的答案的改編如何發出命令來自動創建類似素因式分解的產品？

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\elliptic}{O{i}mmmm}
 {% #1 = item to substitute
  % #2 = main term
  % #3 = first index
  % #4 = last index
  % #5 = operation
  \group_begin:
  \lucas_elliptic:nnnnn { #1 } { #2 } { #3 } { #4 } { #5 }
  \group_end:
 }
\tl_new:N \l__lucas_elliptic_term_tl
\cs_generate_variant:Nn \cs_set:Nn { NV }
\cs_new:Nn \lucas_elliptic:nnnnn
 {
  \tl_set:Nn \l__lucas_elliptic_term_tl { #2 }
  \regex_replace_all:nnN
   { #1 } % search
   { \cB\{\cP\#1\cE\} } % replace
   \l__lucas_elliptic_term_tl % what to act on
  \cs_set:NV \__lucas_elliptic_term:n \l__lucas_elliptic_term_tl
  \__lucas_elliptic_term:n { #3 }
  #5 \dots #5
  \__lucas_elliptic_term:n { #4 }
 }
\ExplSyntaxOff

\begin{document}

$\elliptic{x^{i}}{0}{5}{+}$

$\elliptic{x_{i}}{0}{5}{+}$

$\elliptic[k]{(x_{k}+y_{k}i)}{1}{n}{}$

\end{document}

如果您願意使用#1佔位符，則可以簡化：

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\elliptic}{mmmm}
 {% #1 = main term
  % #2 = first index
  % #3 = last index
  % #4 = operation
  \group_begin:
  \lucas_elliptic:nnnn { #1 } { #2 } { #3 } { #4 }
  \group_end:
 }
\cs_new:Nn \lucas_elliptic:nnnn
 {
  \cs_set:Nn \__lucas_elliptic_term:n { #1 }
  \__lucas_elliptic_term:n { #2 }
  #4 \dots #4
  \__lucas_elliptic_term:n { #3 }
 }
\ExplSyntaxOff

\begin{document}

$\elliptic{x^{#1}}{0}{5}{+}$

$\elliptic{x_{#1}}{0}{5}{+}$

$\elliptic{(x_{#1}+y_{#1}i)}{1}{n}{}$

\end{document}

Question 2

對於參數位於佔位符末尾的範例，您不需要定義任何內部命令，但對於一般情況，該形式\ellipsisb將任何單參數命令定義的主體作為第一個參數。這允許最後顯示 0^2...5^2 形式。

\documentclass{article}



\begin{document}

\newcommand\ellipsis[4]{#1{#2}#4\cdots#4#1{#3}}

$\ellipsis{x^}{0}{5}{+}$

\newcommand\ellipsisb[4]{%
\def\tmp##1{#1}\tmp{#2}#4\cdots#4\tmp{#3}}

$\ellipsisb{x^{#1}}{0}{5}{+}$

$\ellipsisb{{#1}^2}{0}{5}{+}$



\end{document}

Answer

對於參數位於佔位符末尾的範例，您不需要定義任何內部命令，但對於一般情況，該形式\ellipsisb將任何單參數命令定義的主體作為第一個參數。這允許最後顯示 0^2...5^2 形式。

\documentclass{article}



\begin{document}

\newcommand\ellipsis[4]{#1{#2}#4\cdots#4#1{#3}}

$\ellipsis{x^}{0}{5}{+}$

\newcommand\ellipsisb[4]{%
\def\tmp##1{#1}\tmp{#2}#4\cdots#4\tmp{#3}}

$\ellipsisb{x^{#1}}{0}{5}{+}$

$\ellipsisb{{#1}^2}{0}{5}{+}$



\end{document}

Question 3

也許我不明白你想要什麼，但據我所知你不需要\csuse在這裡並且可以定義：

\newcommand\ellip[4]{{#1}^{#3}#2\dots#2{#1}^{#4}}

這樣 $\ellip x+04$ 和 $\ellip y-{-1}2$ 分別產生

如果您確實需要一個支援巨集的更高級版本，那麼我建議不要將巨集放在指數中，而只需替換x^為\csuse{#1}：

\newcommand\fancyellip[4]{\csuse{#1}{#3}#2\dots#2\csuse{#1}{#4}}

這樣現在 $\fancyellip{xint}-{1}2$ 產生

的適當定義\xint。

這是完整的程式碼：

\documentclass{article}

\newcommand\ellip[4]{{#1}^{#3}#2\dots#2{#1}^{#4}}

\usepackage{etoolbox}
\newcommand\fancyellip[4]{\csuse{#1}{#3}#2\dots#2\csuse{#1}{#4}}
\newcommand\xint[1]{\int_{0}^{#1}x\,dx}

\begin{document}

  $\ellip x+04$

  $\ellip y-{-1}2$

  \bigskip

  $\fancyellip{xint}-{1}2$
\end{document}

Answer