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条件中的条件

tag-icon • tag-icon conditionals programming
条件中的条件

考虑以下逻辑陈述：

If \conditionA != 100 and \conditionB != 100 do ``Something~A''
If \conditionA != 100 and \conditionB  = 100 do ``Something~B''
If \conditionA  = 100 and \conditionB != 100 do ``Something~C''
If \conditionA  = 100 and \conditionB  = 100 do ``Something~D''

我如何将其翻译成

\ifnum ... do ... \else do ... \fi

论点类型？

我发现这个帖子由 Joseph Wright 编写，但我不知道如何修改示例以使其实现我想要的效果。

PS 请随意添加缺失的标签。

答案1

如果需要，可以使用 LaTeX3 语法来扩展该命令：

\documentclass{article}
\usepackage{expl3}
\ExplSyntaxOn

\int_new:N \l_svend_condition_A_int
\int_new:N \l_svend_condition_B_int

\cs_new:Npn \svend_check:
 {
  \int_compare:nTF { \l_svend_condition_A_int = 100 }
   {
    \int_compare:nTF { \l_svend_condition_B_int = 100 }
     {
      \typeout{Both~are~100}
     }
     {
      \typeout{A~is~100,~B~is~not~100}
     }
   }
   {
    \int_compare:nTF { \l_svend_condition_B_int = 100 }
     {
      \typeout{A~is~not~100,~B~is~100}
     }
     {
      \typeout{A~is~not~100,~B~is~not~100}
     }
   }
 }

\int_set:Nn \l_svend_condition_A_int {100}
\int_set:Nn \l_svend_condition_B_int {100}
\svend_check:
\int_set:Nn \l_svend_condition_A_int {100}
\int_set:Nn \l_svend_condition_B_int {101}
\svend_check:
\int_set:Nn \l_svend_condition_A_int {101}
\int_set:Nn \l_svend_condition_B_int {100}
\svend_check:
\int_set:Nn \l_svend_condition_A_int {101}
\int_set:Nn \l_svend_condition_B_int {101}
\svend_check:

这是终端上的输出：

Both are 100
A is 100, B is not 100
A is not 100, B is 100
A is not 100, B is not 100

使用传统语法是一样的：

\documentclass{article}

\newcommand{\Check}[2]{%
  \ifnum#1=100
    \ifnum#2=100
      \typeout{\#1 is 100, \#2 is 100}%
    \else
      \typeout{\#1 is 100, \#2 is not 100}%
    \fi
  \else
    \ifnum#2=100
      \typeout{\#1 is not 100, \#2 is 100}%
    \else
      \typeout{\#1 is not 100, \#2 is not 100}%
    \fi
  \fi
}

\Check{100}{100}
\Check{101}{100}
\Check{100}{101}
\Check{101}{101}

输出如下：

\#1 is 100, \#2 is 100
\#1 is not 100, \#2 is 100
\#1 is 100, \#2 is not 100
\#1 is not 100, \#2 is not 100

答案2

为了完整起见，这里有一个etoolbox执行：

在此处输入图片描述

\documentclass{article}
\usepackage{etoolbox}% http://ctan.org/pkg/etoolbox
\newcommand{\Test}[2]{%
  \ifnumequal{#1}{100}{%
    \ifnumequal{#2}{100}{%
      First arg = 100; Second arg = 100%
    }{%
      First arg = 100; Second arg != 100%
    }}{%
    \ifnumequal{#2}{100}{%
      First arg !=100; Second arg = 100%
    }{%
      First arg !=100; Second arg != 100%
    }}%      
}
\begin{document}

\Test{100}{100} \par
\Test{ 99}{100} \par
\Test{101}{101} \par
\Test{100}{101}

\end{document}

答案3

解决方案是否必须达到 TeX 基元级别？如果不是，我想您可以加载ifthen包并执行如下操作：

\ifthenelse{\NOT\conditionA=100}%
   {\ifthenelse{\NOT\ConditionB=100}%
      {"Do SomethingA"}%
      {"Do SomethingB"}}%
   {\ifthenelse{\NOT\ConditionB=100}%
      {"Do SomethingC"}%
      {"Do SomethingD"}}

这里，我假设\ConditionA并\ConditionB求值/扩展为整数。如果不是这种情况，则\NOT...分支将始终占上风。

附录：假设宏\ConditionA和\ConditionB是可扩展的并且求值为整数，那么您也可以使用\ifnum“原始”控制序列。请注意，为了简单起见，测试的是相等而不是不等；因此“Do”序列相对于前面的示例是相反的。

\ifnum\conditionA=100%
   \ifnum\conditionB=100
      {"Do SomethingD"}
      \else {"Do SomethingC"}
   \fi
\else
   \ifnum\conditionB=100
      {"Do SomethingB"}
      \else {"Do SomethingA"}
   \fi
\fi

答案4

为了完整性，(希望很快被完全弃用的) TeX-core 解决方案是完全可扩展的，就像@egreg 的解决方案一样：

\long\def\FirstOfFour#1#2#3#4{#1}
\long\def\SecondOfFour#1#2#3#4{#2}
\long\def\ThirdOfFour#1#2#3#4{#3}
\long\def\FourthOfFour#1#2#3#4{#4}
\def\HundredTest#1#2{%
  \ifcase\numexpr
    \ifnum #1=100 1\else0\fi +
    \ifnum #2=100 2\else0\fi \relax
    \expandafter\FirstOfFour\or
    \expandafter\SecondOfFour\or
    \expandafter\ThirdOfFour\or
    \expandafter\FourthOfFour
  \fi
}

\HundredTest{100}{111}{NN}{YN}{NY}{YY} %YN
\HundredTest{100}{100}{NN}{YN}{NY}{YY} %YY
\HundredTest{111}{111}{NN}{YN}{NY}{YY} %NN
\HundredTest{111}{100}{NN}{YN}{NY}{YY} %NY

\bye

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