I know how to define a color using \definecolor with literals to describe the color. How can I define a color using an already defined color? In other words: How can I make a color accessible under multiple names?
[Tex/LaTex] Defining a Color using another
color
Related Solutions
A simple solution would be to use some predefined gray color and the ! syntax together with \colorlet; for example:
\colorlet{shadecolor}{gray!40}
(40% gray and 60% white); increasing the value used will produce darker shades and decreasing it, lighter ones.
\documentclass{article}
\makeatletter
\def\foo#1\normalcolor\vbox\bgroup#2!!{%
\def\@xfloat ##1[##2]{#1%
\normalcolor
\hbox\bgroup{\color{yellow}\leaders\vrule\hskip\columnwidth\hskip-\columnwidth}%
\vbox \bgroup\aftergroup\egroup
#2}}
\expandafter\foo\@xfloat{#1}[#2]!!
\makeatother
\usepackage{color}
\begin{document}
\begin{figure}
a\\b\\c
\caption{yes no}
\end{figure}
one two three
\end{document}
Best Answer
This can be simply done using
\colorlet{mycolor}{someothercolor}from thexcolorpackage. Heresomeothercolorcan be any other color name or a color mix likecolora!<number>!colorb.You should note that this copies the current color definition. If
someothercoloris changed afterwards (e.g. using another\colorlet)mycolorwill not change.