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定义了一个画立方体的形状，填充颜色的时候出了点问题，虽然解决了但是并不知道原理，MWE如下
```tex
\documentclass[tikz]{standalone}
%% Language and font encodings
\usepackage[english]{babel}
\usepackage[utf8x]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{xcolor}
\usetikzlibrary{positioning}
% Define parallelepiped shape:
\makeatletter
\pgfkeys{/pgf/.cd,
xoffset/.initial=2mm,
yoffset/.initial=2mm
}
\pgfdeclareshape{parallelepiped}
{
\inheritsavedanchors[from=rectangle] % this is nearly a rectangle
\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]{south}
\inheritanchor[from=rectangle]{south west}
\inheritanchor[from=rectangle]{south east}
\backgroundpath{
% store lower right in xa/ya and upper right in xb/yb
\southwest \pgf@xa=\pgf@x \pgf@ya=\pgf@y
\northeast \pgf@xb=\pgf@x \pgf@yb=\pgf@y
\pgfmathsetlength\pgfutil@tempdima{\pgfkeysvalueof{/pgf/xoffset}}
\pgfmathsetlength\pgfutil@tempdimb{\pgfkeysvalueof{/pgf/yoffset}}
\def\ppd@offset{\pgfpoint{\pgfutil@tempdima}{\pgfutil@tempdimb}}
% front rectangle
\pgfpathmoveto{\pgfqpoint{\pgf@xa}{\pgf@ya}}
\pgfpathlineto{\pgfqpoint{\pgf@xb}{\pgf@ya}}
\pgfpathlineto{\pgfqpoint{\pgf@xb}{\pgf@yb}}
\pgfpathlineto{\pgfqpoint{\pgf@xa}{\pgf@yb}}
\pgfpathclose
% other sides
\pgfpathmoveto{\pgfqpoint{\pgf@xb}{\pgf@yb}}
\pgfpathlineto{\pgfpointadd{\pgfpoint{\pgf@xb}{\pgf@yb}}{\ppd@offset}}
\pgfpathlineto{\pgfpointadd{\pgfpoint{\pgf@xa}{\pgf@yb}}{\ppd@offset}}
\pgfpathlineto{\pgfpointadd{\pgfpoint{\pgf@xa}{\pgf@ya}}{\ppd@offset}}
\pgfpathlineto{\pgfqpoint{\pgf@xa}{\pgf@ya}}
\pgfpathmoveto{\pgfqpoint{\pgf@xa}{\pgf@yb}}
\pgfpathlineto{\pgfpointadd{\pgfpoint{\pgf@xa}{\pgf@yb}}{\ppd@offset}}
}
}
\pgfdeclareshape{parallelepiped2}
{
\inheritsavedanchors[from=rectangle] % this is nearly a rectangle
\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]{south}
\inheritanchor[from=rectangle]{south west}
\inheritanchor[from=rectangle]{south east}
\backgroundpath{
% store lower right in xa/ya and upper right in xb/yb
\southwest \pgf@xa=\pgf@x \pgf@ya=\pgf@y
\northeast \pgf@xb=\pgf@x \pgf@yb=\pgf@y
\pgfmathsetlength\pgfutil@tempdima{\pgfkeysvalueof{/pgf/xoffset}}
\pgfmathsetlength\pgfutil@tempdimb{\pgfkeysvalueof{/pgf/yoffset}}
\def\ppd@offset{\pgfpoint{\pgfutil@tempdima}{\pgfutil@tempdimb}}
\pgfpathmoveto{\pgfqpoint{\pgf@xa}{\pgf@ya}}
\pgfpathlineto{\pgfqpoint{\pgf@xb}{\pgf@ya}}
\pgfpathlineto{\pgfqpoint{\pgf@xb}{\pgf@yb}}
\pgfpathlineto{\pgfqpoint{\pgf@xa}{\pgf@yb}}
\pgfpathclose
\pgfpathmoveto{\pgfqpoint{\pgf@xa}{\pgf@ya}}
\pgfpathlineto{\pgfpointadd{\pgfpoint{\pgf@xa}{\pgf@ya}}{\ppd@offset}}
\pgfpathlineto{\pgfpointadd{\pgfpoint{\pgf@xa}{\pgf@yb}}{\ppd@offset}}
\pgfpathlineto{\pgfpointadd{\pgfpoint{\pgf@xb}{\pgf@yb}}{\ppd@offset}}
\pgfpathlineto{\pgfqpoint{\pgf@xb}{\pgf@yb}}
\pgfpathmoveto{\pgfqpoint{\pgf@xa}{\pgf@yb}}
\pgfpathlineto{\pgfpointadd{\pgfpoint{\pgf@xa}{\pgf@yb}}{\ppd@offset}}
}
}
\makeatother
\tikzset{
conv/.style={
parallelepiped, draw,
xoffset=-0.5cm,
yoffset=0.5cm,
path picture={
\draw[top color=red!50,bottom color=red!50]
(path picture bounding box.south west) rectangle
(path picture bounding box.north east);
},
text=black,
},
conv2/.style={
parallelepiped2, draw,
xoffset=-0.5cm,
yoffset=0.5cm,
path picture={
\draw[top color=red!50,bottom color=red!50]
(path picture bounding box.south west) rectangle
(path picture bounding box.north east);
},
text=black,
},
}
\begin{document}
\begin{tikzpicture}
\node[conv, minimum width=3cm, minimum height=1cm](conv1){Conv};
\node[conv2, right=1cm of conv1, minimum width=3cm, minimum
height=1cm](conv1){Conv2};
\end{tikzpicture}
\end{document}
```
运行结果如下
2 回答
0
如果试试
```tex
\tikz{
\filldraw [fill=red!50]
(0,0)--(1,0)--(1,1)--(0,1)--cycle
(0,0)--($(0,0)+(135:0.5)$)--($(0,1)+(135:0.5)$)--($(1,1)+(135:0.5)$)--(1,1)
(0,1)--($(0,1)+(135:0.5)$);
}
```
就看到类似的问题，当填充路径时，非封闭的
``(0,0)--($(0,0)+(135:0.5)$)--($(0,1)+(135:0.5)$)--($(1,1)+(135:0.5)$)--(1,1)``
会被自动封闭，即首尾连接起来，按非零规则，白色区域不属于被填充区域。试试
```
\tikz{
\filldraw [fill=red!50]
(0,0)--(1,0)--(1,1)--(0,1)--cycle
(0,0)--($(0,0)+(135:0.5)$)--($(0,1)+(135:0.5)$)--($(1,1)+(135:0.5)$)--(1,1)--(0,1)--cycle
(0,1)--($(0,1)+(135:0.5)$);
}
```
或者
```
\tikz{
\filldraw [fill=red!50]
(0,0)--(0,1)--(1,1)--(1,0)--cycle
(0,0)--($(0,0)+(135:0.5)$)--($(0,1)+(135:0.5)$)--($(1,1)+(135:0.5)$)--(1,1)
(0,1)--($(0,1)+(135:0.5)$);
}
```
就没有这种问题。
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