4

cycle doesn't work as I would naïvely expect it to if I use the node operation rather than coordinate to mark my points. I realise that, in the case of a node, TikZ is drawing to and from the borders of the nodes. So I can understand that the triangle is not filled because TikZ might regard the path as a non-closed one. (Although in other cases, it will actually finish a path in order to fill it as requested.) But why is the final size of the triangle not even drawn?

\documentclass[tikz,border=10pt,multi]{standalone}
\begin{document}
\begin{tikzpicture}[every node/.append style={draw}]
  \node (a)  at (-1,1) {};
  \node (c)  at (-1,-1) {};
  \node (f)  at (1,1) {};
  \filldraw (a) -- (c) -- (f) -- cycle;
\end{tikzpicture}
\end{document}

pmav99 asked this question in 2011. However, the answers there do not obviously explain to me why nothing is drawn at all. I could understand if the final line returned to the border anchor of (a) from which the path began. I could also understand if it returned to the border anchor of (a) which it would use if I wrote -- (a) rather than -- cycle. But why does it not appear to go anywhere? (Or where does it go, if it goes somewhere, and why?)

I'm sure this must be a duplicate, but I can't find an explanation of this specific aspect of what TikZ is doing.

Also, I of course know that \coordinate (a) ... etc. would make more sense here, that I could define the points while constructing the path etc. I know how to avoid the problem. What I want to understand is the precise whys of why that problem occurs in the particular form it does. What exactly is -- cycle doing in this case?

6

Because each of them are separate paths from each node shape border to the other.

The final construction is equivalent to (using convenience of their locations)

\filldraw (a.-90) -- (c.90) % pen is lifted here hence a new path starts
          (c.45) -- (f.-135) -- cycle;

Hence cycle goes back to (c.45) as it is the starting point of the unbroken path redrawing the last portion of the path creating a zero area.

Put anything that is not collinear between (c) and (f) and you get a fill

\filldraw (a) -- (c) -- (1,-1) --(f) -- cycle;

enter image description here

Coordinates on the other hand have no border shapes but only a center anchor hence they constitute as valid path points thus cycle goes back to a since that is the first point of the continuous path.

  • Nitpick: But it is one whole path, not separate ones. But yes, a move to operation happens. And the PGFmanual says: “This causes the straight line from the current point to go to the last point specified by a move-to operation.” – Qrrbrbirlbel Jun 19 '16 at 10:52
  • @Qrrbrbirlbel Yes probably I should have said pen down instead unbroken since it is a PDF operation. – percusse Jun 19 '16 at 11:14

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