
我有以下宏
\usepackage{xintexpr}
\newcommand\seq[4] %length, delimiter, generating func, last term
{%
\def\s##1{\def\n{##1} #3}
\xintListWithSep{#2}
{%
\xintApply{\s}{\xintSeq{1}{#1}}
}%
#2\ldots
\ifx\\#4\\ %if 4th arg empty
%empty
\else
#2#4
\fi
}
該類型使用給定的生成函數來設定一個序列,即
$\seq{3}{,}{\n}{n}$
$\seq{4}{,}{\n}{}$
$\seq{3}{,}{\sqrt{\n}}{}$
$\seq{3}{/}{\frac{1}{\n}}{\frac{1}{n}}$
相當於
$1, 2, 3, ..., n$
$1, 2, 3, 4, ...$
$\sqrt{1}, \sqrt{2}, \sqrt{3}, ...$
$\frac{1}{1}/ \frac{1}{2}/ \frac{1}{3}/ .../ \frac{1}{n}$
儘管它按預期工作,但當它處於支援對齊的環境中並傳遞對齊製表符 ( &
) 作為第二個參數且第四個參數為空時,它會失敗,即
\begin{align}
\seq{3}{&&,}{\n}{n} % this works
\seq{3}{&&,}{\n}{} % this causes an error
\end{align}
在修改程式碼後,我得出的結論是,\ifx
僅當評估為%empty
且前面有字元時,該錯誤才由該語句引起,如以下範例所示:
\documentclass{article}
\usepackage{amsmath}
\newcommand\testa[2]
{%
\ifx\\#2\\
%empty, evaluation causes error
\else
#1
\fi
}
\newcommand\testb[2]
{%
\ifx\\#2\\
#1
\else
%full, evaluation causes error
\fi
}
\begin{document}
\begin{align}
\testa{&}{n} \\ %ok
\testb{&}{} \\ %ok
%
\testa{&}{} \\ %ok
\testb{&}{n} \\ %ok
%
foo\testa{&}{n} \\ %ok
foo\testb{&}{} \\ %ok
%
foo\testa{&}{} \\ % causes error
foo\testb{&}{n} % causes error
\end{align}
\end{document}
如何更改條件來處理對齊製表符?
如果問題很具體,請告訴我,我會將其刪除。
編輯:加入\usepackage{amsmath}
到範例中
答案1
您需要隱藏&
正在跳過的分支
\documentclass{article}
\usepackage{amsmath}
\def\useone#1{#1}
\newcommand\testa[2]
{%
\ifx\\#2\\%%
%empty, evaluation causes error
\else
\useone{#1}%%
\fi
}
\newcommand\testb[2]
{%
\ifx\\#2\\%%
\useone{#1}%%
\else
%full, evaluation causes error
\fi
}
\begin{document}
\begin{align}
\testa{&}{n} \\ %ok
\testb{&}{} \\ %ok
%
\testa{&}{} \\ %ok
\testb{&}{n} \\ %ok
%
foo\testa{&}{n} \\ %ok
foo\testb{&}{} \\ %ok
%
foo\testa{&}{} \\ % causes error
foo\testb{&}{n} % causes error
\end{align}
\end{document}