累积上标运算符：将 $^i^+^1$ 转换为 $^{i+1}$ 而不是双上标 ${{^i}^{+}+1}

Question 1

宏\SUP应该定义：

\def\SUP#1{\def\SUPt{#1}\futurelet\next\SUPa}
\def\SUPa{\ifx\next\SUP^\bgroup\SUPt\let\SUP=\SUPb\else^{\SUPt}\fi}
\def\SUPb#1{#1\futurelet\next\SUPc}
\def\SUPc{\ifx\next\SUP\else\egroup\fi}

%test:
$a\SUP2+b\SUP2=c\SUP2$

$a\SUP1\SUP2\SUP3$

\bye

Answer

宏\SUP应该定义：

\def\SUP#1{\def\SUPt{#1}\futurelet\next\SUPa}
\def\SUPa{\ifx\next\SUP^\bgroup\SUPt\let\SUP=\SUPb\else^{\SUPt}\fi}
\def\SUPb#1{#1\futurelet\next\SUPc}
\def\SUPc{\ifx\next\SUP\else\egroup\fi}

%test:
$a\SUP2+b\SUP2=c\SUP2$

$a\SUP1\SUP2\SUP3$

\bye

Question 2

这适用于 XeLaTeX 或 LuaLaTeX：

\documentclass{article}
\usepackage{newunicodechar}

\ExplSyntaxOn

\cs_new_protected:Nn \yossi_sup:
 {
  \peek_regex_replace_once:nn
   { ([¹²³⁴⁵⁶⁷⁸⁹⁰]*) }
   { \c{sp}\cB\{\c{yossi_replace:n}\cB\{\1\cE\}\cE\} }
 }
\cs_new_protected:Nn \yossi_replace:n
 {
  \tl_map_function:nN { #1 } \__yossi_replace:n
 }
\cs_new_protected:Nn \__yossi_replace:n
 {
  \str_case:nn { #1 } { {¹}{1}{²}{2}{³}{3}{⁴}{4}{⁵}{5}{⁶}{6}{⁷}{7}{⁸}{8}{⁹}{9}{⁰}{0} }
 }
\newunicodechar{¹}{\yossi_sup: ¹}
\newunicodechar{²}{\yossi_sup: ²}
\newunicodechar{³}{\yossi_sup: ³}
\newunicodechar{⁴}{\yossi_sup: ⁴}
\newunicodechar{⁵}{\yossi_sup: ⁵}
\newunicodechar{⁶}{\yossi_sup: ⁶}
\newunicodechar{⁷}{\yossi_sup: ⁷}
\newunicodechar{⁸}{\yossi_sup: ⁸}
\newunicodechar{⁹}{\yossi_sup: ⁹}
\newunicodechar{⁰}{\yossi_sup: ⁰}

\ExplSyntaxOff

\begin{document}

$a²³ + b²³ = b²³$

$a¹²³⁴⁵⁶⁷⁸⁹⁰$

\end{document}

每个上标数字都有一个以开头的定义。这将开始工作：收集\yossi_sup:列表中的每个后续字符（包括起始字符），直到列表中没有其他字符。之后，将所有内容放入其中，并用标准对应字符逐一替换字符。¹²³⁴⁵⁶⁷⁸⁹⁰^{...}

它可能对于也是可行的pdflatex，但主要问题是每个上标数字都是两个或三个字节长，最终的替换会复杂得多。

Answer

这适用于 XeLaTeX 或 LuaLaTeX：

\documentclass{article}
\usepackage{newunicodechar}

\ExplSyntaxOn

\cs_new_protected:Nn \yossi_sup:
 {
  \peek_regex_replace_once:nn
   { ([¹²³⁴⁵⁶⁷⁸⁹⁰]*) }
   { \c{sp}\cB\{\c{yossi_replace:n}\cB\{\1\cE\}\cE\} }
 }
\cs_new_protected:Nn \yossi_replace:n
 {
  \tl_map_function:nN { #1 } \__yossi_replace:n
 }
\cs_new_protected:Nn \__yossi_replace:n
 {
  \str_case:nn { #1 } { {¹}{1}{²}{2}{³}{3}{⁴}{4}{⁵}{5}{⁶}{6}{⁷}{7}{⁸}{8}{⁹}{9}{⁰}{0} }
 }
\newunicodechar{¹}{\yossi_sup: ¹}
\newunicodechar{²}{\yossi_sup: ²}
\newunicodechar{³}{\yossi_sup: ³}
\newunicodechar{⁴}{\yossi_sup: ⁴}
\newunicodechar{⁵}{\yossi_sup: ⁵}
\newunicodechar{⁶}{\yossi_sup: ⁶}
\newunicodechar{⁷}{\yossi_sup: ⁷}
\newunicodechar{⁸}{\yossi_sup: ⁸}
\newunicodechar{⁹}{\yossi_sup: ⁹}
\newunicodechar{⁰}{\yossi_sup: ⁰}

\ExplSyntaxOff

\begin{document}

$a²³ + b²³ = b²³$

$a¹²³⁴⁵⁶⁷⁸⁹⁰$

\end{document}

每个上标数字都有一个以开头的定义。这将开始工作：收集\yossi_sup:列表中的每个后续字符（包括起始字符），直到列表中没有其他字符。之后，将所有内容放入其中，并用标准对应字符逐一替换字符。¹²³⁴⁵⁶⁷⁸⁹⁰^{...}

它可能对于也是可行的pdflatex，但主要问题是每个上标数字都是两个或三个字节长，最终的替换会复杂得多。

Question 3

与 wipet 类似，但增加了额外的扩展，因此\SUP可以隐藏在 unicode 字符的扩展后面。

\documentclass{article}

%\usepackage[utf8]{inputenc}% only needed in old latex releases

\def\SUP#1{^\bgroup
#1\expandafter\futurelet\expandafter\next\expandafter\SUPb
\romannumeral`\^^@}
\def\SUPb{\ifx\next^\expandafter\SUPc\else\egroup\fi}
\def\SUPc#1#2{}

\DeclareUnicodeCharacter{00B2}{\SUP2}\DeclareUnicodeCharacter{00B3}{\SUP3}
\begin{document}
I can ask for $a$ squared plus $b$ squared to be equal to $c$ squared, 
  \[a²+b²=c²,\] 
and I can also ask for $a$ cubed plus $b$ cubed to be equal to $c$ cubed, 
  \[a³+b³= c³.\]
But, I cannot ask for $a$ raised to the twenty-third degree plus $b$ raised 
raised to the twenty-third degree to be equal to $c$ raised to the twenty-third 
degree, \[ a²³ + b²³ = b²³. \]
\end{document}

Answer

与 wipet 类似，但增加了额外的扩展，因此\SUP可以隐藏在 unicode 字符的扩展后面。

\documentclass{article}

%\usepackage[utf8]{inputenc}% only needed in old latex releases

\def\SUP#1{^\bgroup
#1\expandafter\futurelet\expandafter\next\expandafter\SUPb
\romannumeral`\^^@}
\def\SUPb{\ifx\next^\expandafter\SUPc\else\egroup\fi}
\def\SUPc#1#2{}

\DeclareUnicodeCharacter{00B2}{\SUP2}\DeclareUnicodeCharacter{00B3}{\SUP3}
\begin{document}
I can ask for $a$ squared plus $b$ squared to be equal to $c$ squared, 
  \[a²+b²=c²,\] 
and I can also ask for $a$ cubed plus $b$ cubed to be equal to $c$ cubed, 
  \[a³+b³= c³.\]
But, I cannot ask for $a$ raised to the twenty-third degree plus $b$ raised 
raised to the twenty-third degree to be equal to $c$ raised to the twenty-third 
degree, \[ a²³ + b²³ = b²³. \]
\end{document}

Question 4

好吧，这基本上是复制粘贴的大卫·卡莱尔的回答--- 我刚刚浏览过https://unicode-table.com/en/blocks/superscripts-and-subscripts/，修改了（我希望是正确的）\romannumeral黑魔法并将其扩展为下标。

我不敢编辑大卫的答案......;-)

这并不能保证它不会吃掉你的小猫——我真的不知道黑魔法的部分格罗克。

\documentclass{article}

%
% Shamelessly copied from  @David Carlisle
% See https://tex.stackexchange.com/a/586136/38080
%

%\usepackage[utf8]{inputenc}% only needed in old latex releases

\def\SUP#1{^\bgroup
#1\expandafter\futurelet\expandafter\next\expandafter\SUPb
\romannumeral`\^^@}
\def\SUPb{\ifx\next^\expandafter\SUPc\else\egroup\fi}
\def\SUPc#1#2{}
\def\SUB#1{_\bgroup
#1\expandafter\futurelet\expandafter\next\expandafter\SUBb
\romannumeral`\^^@}
\def\SUBb{\ifx\next_\expandafter\SUBc\else\egroup\fi}
\def\SUBc#1#2{}

\DeclareUnicodeCharacter{2070}{\SUP0}
\DeclareUnicodeCharacter{00B9}{\SUP1}
\DeclareUnicodeCharacter{00B2}{\SUP2}
\DeclareUnicodeCharacter{00B3}{\SUP3}
\DeclareUnicodeCharacter{2074}{\SUP4}
\DeclareUnicodeCharacter{2075}{\SUP5}
\DeclareUnicodeCharacter{2076}{\SUP6}
\DeclareUnicodeCharacter{2077}{\SUP7}
\DeclareUnicodeCharacter{2078}{\SUP8}
\DeclareUnicodeCharacter{2079}{\SUP9}
\DeclareUnicodeCharacter{2071}{\SUP{i}}
\DeclareUnicodeCharacter{207A}{\SUP{{+}}}
\DeclareUnicodeCharacter{207B}{\SUP{{-}}}
\DeclareUnicodeCharacter{207C}{\SUP{{=}}}
\DeclareUnicodeCharacter{207D}{\SUP{(}}
\DeclareUnicodeCharacter{207E}{\SUP{)}}
\DeclareUnicodeCharacter{207F}{\SUP{n}}
\DeclareUnicodeCharacter{2080}{\SUB0}
\DeclareUnicodeCharacter{2081}{\SUB1}
\DeclareUnicodeCharacter{2082}{\SUB2}
\DeclareUnicodeCharacter{2083}{\SUB3}
\DeclareUnicodeCharacter{2084}{\SUB4}
\DeclareUnicodeCharacter{2085}{\SUB5}
\DeclareUnicodeCharacter{2086}{\SUB6}
\DeclareUnicodeCharacter{2087}{\SUB7}
\DeclareUnicodeCharacter{2088}{\SUB8}
\DeclareUnicodeCharacter{2089}{\SUB9}
\DeclareUnicodeCharacter{208A}{\SUB{{+}}}
\DeclareUnicodeCharacter{208B}{\SUB{{-}}}
\DeclareUnicodeCharacter{208C}{\SUB{{=}}}
\DeclareUnicodeCharacter{208D}{\SUB{(}}
\DeclareUnicodeCharacter{208E}{\SUB{)}}
\DeclareUnicodeCharacter{2090}{\SUB{a}}
\DeclareUnicodeCharacter{2091}{\SUB{e}}
\DeclareUnicodeCharacter{2092}{\SUB{o}}
\DeclareUnicodeCharacter{2093}{\SUB{x}}

\begin{document}
I can ask for $a$ squared plus $b$ squared to be equal to $c$ squared,
  \[a²+b²=c²,\]
and I can also ask for $a$ cubed plus $b$ cubed to be equal to $c$ cubed,
  \[a³+b³= c³.\]
But, I cannot ask for $a$ raised to the twenty-third degree plus $b$ raised
raised to the twenty-third degree to be equal to $c$ raised to the twenty-third
degree, \[ a²³ + b²³ = b²³. \]
\[ x¹²³⁴⁵⁶⁷⁸⁹⁰pⁱⁿv⁺⁻z⁼k⁽¹⁾ \]
\[a¹₀₁₍₃₊₎\]
\end{document}

（再次对糟糕的抗锯齿效果表示抱歉）

Answer

好吧，这基本上是复制粘贴的大卫·卡莱尔的回答--- 我刚刚浏览过https://unicode-table.com/en/blocks/superscripts-and-subscripts/，修改了（我希望是正确的）\romannumeral黑魔法并将其扩展为下标。

我不敢编辑大卫的答案......;-)

这并不能保证它不会吃掉你的小猫——我真的不知道黑魔法的部分格罗克。

\documentclass{article}

%
% Shamelessly copied from  @David Carlisle
% See https://tex.stackexchange.com/a/586136/38080
%

%\usepackage[utf8]{inputenc}% only needed in old latex releases

\def\SUP#1{^\bgroup
#1\expandafter\futurelet\expandafter\next\expandafter\SUPb
\romannumeral`\^^@}
\def\SUPb{\ifx\next^\expandafter\SUPc\else\egroup\fi}
\def\SUPc#1#2{}
\def\SUB#1{_\bgroup
#1\expandafter\futurelet\expandafter\next\expandafter\SUBb
\romannumeral`\^^@}
\def\SUBb{\ifx\next_\expandafter\SUBc\else\egroup\fi}
\def\SUBc#1#2{}

\DeclareUnicodeCharacter{2070}{\SUP0}
\DeclareUnicodeCharacter{00B9}{\SUP1}
\DeclareUnicodeCharacter{00B2}{\SUP2}
\DeclareUnicodeCharacter{00B3}{\SUP3}
\DeclareUnicodeCharacter{2074}{\SUP4}
\DeclareUnicodeCharacter{2075}{\SUP5}
\DeclareUnicodeCharacter{2076}{\SUP6}
\DeclareUnicodeCharacter{2077}{\SUP7}
\DeclareUnicodeCharacter{2078}{\SUP8}
\DeclareUnicodeCharacter{2079}{\SUP9}
\DeclareUnicodeCharacter{2071}{\SUP{i}}
\DeclareUnicodeCharacter{207A}{\SUP{{+}}}
\DeclareUnicodeCharacter{207B}{\SUP{{-}}}
\DeclareUnicodeCharacter{207C}{\SUP{{=}}}
\DeclareUnicodeCharacter{207D}{\SUP{(}}
\DeclareUnicodeCharacter{207E}{\SUP{)}}
\DeclareUnicodeCharacter{207F}{\SUP{n}}
\DeclareUnicodeCharacter{2080}{\SUB0}
\DeclareUnicodeCharacter{2081}{\SUB1}
\DeclareUnicodeCharacter{2082}{\SUB2}
\DeclareUnicodeCharacter{2083}{\SUB3}
\DeclareUnicodeCharacter{2084}{\SUB4}
\DeclareUnicodeCharacter{2085}{\SUB5}
\DeclareUnicodeCharacter{2086}{\SUB6}
\DeclareUnicodeCharacter{2087}{\SUB7}
\DeclareUnicodeCharacter{2088}{\SUB8}
\DeclareUnicodeCharacter{2089}{\SUB9}
\DeclareUnicodeCharacter{208A}{\SUB{{+}}}
\DeclareUnicodeCharacter{208B}{\SUB{{-}}}
\DeclareUnicodeCharacter{208C}{\SUB{{=}}}
\DeclareUnicodeCharacter{208D}{\SUB{(}}
\DeclareUnicodeCharacter{208E}{\SUB{)}}
\DeclareUnicodeCharacter{2090}{\SUB{a}}
\DeclareUnicodeCharacter{2091}{\SUB{e}}
\DeclareUnicodeCharacter{2092}{\SUB{o}}
\DeclareUnicodeCharacter{2093}{\SUB{x}}

\begin{document}
I can ask for $a$ squared plus $b$ squared to be equal to $c$ squared,
  \[a²+b²=c²,\]
and I can also ask for $a$ cubed plus $b$ cubed to be equal to $c$ cubed,
  \[a³+b³= c³.\]
But, I cannot ask for $a$ raised to the twenty-third degree plus $b$ raised
raised to the twenty-third degree to be equal to $c$ raised to the twenty-third
degree, \[ a²³ + b²³ = b²³. \]
\[ x¹²³⁴⁵⁶⁷⁸⁹⁰pⁱⁿv⁺⁻z⁼k⁽¹⁾ \]
\[a¹₀₁₍₃₊₎\]
\end{document}

（再次对糟糕的抗锯齿效果表示抱歉）

累积上标运算符：将 $^i^+^1$ 转换为 $^{i+1}$ 而不是双上标 ${{^i}^{+}+1}

答案1

答案2

答案3

答案4

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