While writing a to operator for TikZ I struck upon the following problem, how does one assign a value to \pgf@x or it's relatives \pgf@xa, \pgf@xb and \pgf@xc ?
Originally I expected something like
...
% Determine the center of the chord connecting the co-ordinates
\pgfmathparse{#1*sin(\tikz@angle@c)}%
\pgf@xc=\pgfmathresult%
\pgfmathparse{#1*cos(\tikz@angle@c)}%
\pgf@yc=\pgfmathresult%
...
to work but this and a number of variations I tried using \let and even \relax failed to assign correctly.
In the, working, code below I have used \pgfsetmacro to assign the temporary values \ctr@x and \ctr@y which I then re-assign to \pgf@x and \pgf@y respectively.
\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\makeatletter
\tikzset{%
arc over/.style={
to path={%
\pgfextra{%
% Retrieve and assign the source co-ordinate
\tikz@scan@one@point\pgf@process(\tikztostart)%
\pgf@xa=\pgf@x\pgf@ya=\pgf@y%
% Retrieve and assign the target co-ordinate
\tikz@scan@one@point\pgf@process(\tikztotarget)%
\pgf@xb=\pgf@x\pgf@yb=\pgf@y
% Determine the slope of the chord connecting the co-ordinates
% \pgfmathanglebetweenpoints{\tikztostart}{\tikztotarget}% This gave funny results
\advance\pgf@x by-\pgf@xa%
\advance\pgf@y by-\pgf@ya%
\pgfmathatantwo{\the\pgf@y}{\the\pgf@x}%
% \pgfmathparse{\pgfmathresult + pi/2}% This behaves wierdly
\pgfmathMod@{\pgfmathresult}{360}%
\pgfmathparse{\pgfmathresult - 90}% Perhaps one should account for sign e.g. +/- 90.
\let\tikz@angle@c=\pgfmathresult%
% Determine the center of the chord connecting the co-ordinates
\pgfmathsetmacro{\ctr@x}{(\pgf@xa+\pgf@xb)/2}
\pgfmathsetmacro{\ctr@y}{(\pgf@ya+\pgf@yb)/2}
\pgf@xc=-\ctr@x\pgf@yc=-\ctr@y%
% Offset the center by the assigned amount
\pgfmathsetmacro{\ctr@x}{#1*sin(\tikz@angle@c)}
\pgfmathsetmacro{\ctr@y}{#1*cos(\tikz@angle@c)}
\advance\pgf@xc by\ctr@x pt%
\advance\pgf@yc by\ctr@y pt%
% Normalize the co-ordinates
\advance\pgf@xa by-\pgf@xc%
\advance\pgf@ya by-\pgf@yc%
\advance\pgf@xb by-\pgf@xc%
\advance\pgf@yb by-\pgf@yc%
% Determine the start and end angles
\pgfmathatantwo{\the\pgf@ya}{\the\pgf@xa}%
\pgfmathMod@{\pgfmathresult}{360}%
\let\tikz@angle@a=\pgfmathresult%
\pgfmathatantwo{\the\pgf@yb}{\the\pgf@xb}%
\pgfmathMod@{\pgfmathresult}{360}%
\let\tikz@angle@b=\pgfmathresult%
% Determine the radius of the arc to be drawn
\pgfmathveclen{\pgf@xa}{\pgf@ya}%
\let\tikz@radius=\pgfmathresult%
% Define the arc that is to be drawn
\edef\tikz@to@arc@path{ arc(\tikz@angle@a:\tikz@angle@b:\tikz@radius pt) }%\show\tikz@to@arc@path
}%
\tikz@to@arc@path
}
}
}
\makeatother
\begin{document}
\begin{tikzpicture}
\draw (9pt,0) to[arc over= 0.5em] (0,9pt)
(9pt,0) -- (0,9pt)
( 0,0) circle (10 pt);
\end{tikzpicture}
\end{document}
This seems like bad form and I was wondering if there is a better means of achieving this.
Co-incidently if one feels like providing a code review of sorts I would be most grateful, I'm not entirely familiar with the lower level TeX stuff and this is hopefully a small example that I might learn from.
My code is adapted from the answers of Mark Wibrows and Loop Space. This question is somewhat related but did not resolve my problem.
\pgfmathparse? – percusse Jan 13 '17 at 23:22\pgfmathresultrather then\pgfmathparse{...}, the former results when one executes the latter. The trouble is I don't seem to be able to do the assignment e.g. should one use\pgf@x = \pgfmathresultor\let\pgf@x = \the\pgfmathresult? I think I don't really understand why this is, from other programming experience this feels like I'm mismatching types i.e. Am I assigning a\pgfmath{#1*sin(\tikz@angle@c)}%– percusse Jan 15 '17 at 14:01\pgfmathsetlength? – Symbol 1 Jan 28 '17 at 00:50