为块模式图定义 tikz 形状

Question

我想做的事

我想绘制分组密码操作模式的图表，例如这个ECB 模式。我以前这样做过，但没有为图表的特定部分定义形状。但是，这次我想正确地做到这一点，并为 XOR 运算和明文、密文和 IV 的块定义形状。

对于 XOR 运算这个问题之前已经问过。但是，我需要定义一个新的形状，即明文、密文和 IV 的块。

我的方法

我现在的方法是基于矩形形状定义一个新形状，并使用 for 循环在矩形内绘制 9 条均匀分布的垂直线。这样，矩形内就会产生 10 个大小相同的隔间。为此，我计算了两条垂直线之间的距离。然后，我从保存的锚点开始north.east，在计算出的距离内绘制 9 条垂直线。我在以下 MWE 中执行了此操作：

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}

\makeatletter
\pgfdeclareshape{block}
{
  \inheritsavedanchors[from=rectangle] % this is a rectanble
  \inheritanchorborder[from=rectangle]
  \inheritanchor[from=rectangle]{north}
  \inheritanchor[from=rectangle]{north west}
  \inheritanchor[from=rectangle]{north east}
  \inheritanchor[from=rectangle]{center}
  \inheritanchor[from=rectangle]{west}
  \inheritanchor[from=rectangle]{east}
  \inheritanchor[from=rectangle]{mid}
  \inheritanchor[from=rectangle]{mid west}
  \inheritanchor[from=rectangle]{mid east}
  \inheritanchor[from=rectangle]{base}
  \inheritanchor[from=rectangle]{base west}
  \inheritanchor[from=rectangle]{base east}
  \inheritanchor[from=rectangle]{south}
  \inheritanchor[from=rectangle]{south west}
  \inheritanchor[from=rectangle]{south east}
  \inheritbackgroundpath[from=rectangle]

  \foregroundpath{
    \setlength{\pgf@ya}{\pgfshapeminheight}
    \setlength{\pgf@xa}{\pgfshapeminwidth}
    \pgfmathsetlength\pgfutil@tempdima{\pgf@xa / 10}
    \pgfpathmoveto{\northeast}
    \foreach \i in {1,...,9}
    {%
      \pgfpathmoveto{\pgfpointadd{\pgfpoint{\pgf@x}{\pgf@y}}{\pgfpoint{-\pgfutil@tempdima}{0pt}}}
      \pgfpathlineto{\pgfpointadd{\pgfpoint{\pgf@x}{\pgf@y}}{\pgfpoint{0pt}{-2\pgf@ya}}}
    }
  }
}

\makeatother


\begin{document}
\tikzset{blockstyle/.style={draw, minimum width = 4cm, minimum height = 0.4cm, shape=block}}
\begin{tikzpicture}
\node[blockstyle] (p1) at (0.1cm,0.1cm)  {};
\node[above] at (p1.north) {Plaintext 1};x
\end{tikzpicture}

\end{document}

问题

在 处绘制一个块(0,0)可以正常工作，但是一旦在另一个位置绘制一个块，垂直线的绘制就会与矩形外绘制的线条混淆，如下图所示：

因此我的问题是：绘制线条的计算有什么问题？

编辑似乎问题发生在节点未放置在 时(0,0)。这可能是问题根源的线索。我采用了 MWE 和相应的文本。

Answer 1

我把循环改得更明确一些，这似乎有效。你说的对\pgfpathmoveto{\northeast}。我还给形状和节点样式赋予了不同的名称。（我还要提一下，你也可以使用pics 和local bounding box）。

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}

\makeatletter
\pgfdeclareshape{fblock}
{
  \inheritsavedanchors[from=rectangle] % this is a rectanble
  \inheritanchorborder[from=rectangle]
  \inheritanchor[from=rectangle]{north}
  \inheritanchor[from=rectangle]{north west}
  \inheritanchor[from=rectangle]{north east}
  \inheritanchor[from=rectangle]{center}
  \inheritanchor[from=rectangle]{west}
  \inheritanchor[from=rectangle]{east}
  \inheritanchor[from=rectangle]{mid}
  \inheritanchor[from=rectangle]{mid west}
  \inheritanchor[from=rectangle]{mid east}
  \inheritanchor[from=rectangle]{base}
  \inheritanchor[from=rectangle]{base west}
  \inheritanchor[from=rectangle]{base east}
  \inheritanchor[from=rectangle]{south}
  \inheritanchor[from=rectangle]{south west}
  \inheritanchor[from=rectangle]{south east}
  \inheritbackgroundpath[from=rectangle]

  \foregroundpath{
    \setlength{\pgf@ya}{\pgfshapeminheight}
    \setlength{\pgf@xa}{\pgfshapeminwidth}
    \pgfmathsetlength\pgfutil@tempdima{\pgf@xa / 10}
    \foreach \i in {1,...,9}
    {  
      \pgfpathmoveto{\pgfpointadd{\northeast}{\pgfpoint{-\i*\pgfutil@tempdima}{0pt}}}
      \pgfpathlineto{\pgfpointadd{\northeast}{\pgfpoint{-\i*\pgfutil@tempdima}{-2\pgf@ya}}}
    }
  }
}

\makeatother


\begin{document}
\tikzset{block/.style={draw, minimum width = 4cm, minimum height = 0.4cm,
shape=fblock}}
\begin{tikzpicture}%[every block/.style={draw}]
\node[block] (p1) at (0,0)  {};
\node[above] at (p1.north) {Plaintext 1};
\node[block, left=1.5cm] (iv) at (p1.west) {};
\end{tikzpicture}
\end{document}