Cutting out shape to reveal what's behind

Question

The following MWE displays the desired output, but the question is whether (and how) the same result can be achieved by somehow redefining lines (1) and (2) to "cut out" the inner square, so that line (3) is redundant. I appreciate that this picture is simple, but one can easily imagine more complicated cases.

\documentclass{article}
\usepackage{tikz}

\begin{document}

\begin{tikzpicture}
\draw (-1,-1)--++(4,4);
\draw[fill=white] (0,0)--++(2,0)--++(0,2)--++(-2,0)--++(0,-2); % (1)
\draw (0.5,0.5)--++(1,0)--++(0,1)--++(-1,0)--++(0,-1); % (2)
\draw (0.5,0.5)--++(1,1); % (3)
\end{tikzpicture}

\end{document}

Answer 1

Filling

If you actually want to fill the outer rectangle but not the inner, you can use the even odd rule (PGF manual, pp. 163f.)

Code

\documentclass[tikz]{standalone}
\begin{document}
\begin{tikzpicture}[even odd rule]
\draw (-1,-1) -- ++ (4,4);
\draw[fill=white] (0,0) rectangle ++ (2,2) (.5,.5) rectangle ++ (1,1);
\end{tikzpicture}
\end{document}

Output

Clipping

Another approach would be clipping, but as you want to “reverse-clip” the line to the boxes you will need a tweaked version of clip.

In this small example the following code suffices, but if you want to draw more or more sophisticated pictures you will need the reverseclip style from the referenced answer and/or a scope.

Reference

• Jake’s answer to How can I invert a 'clip' selection within TikZ?

Code

\documentclass[tikz]{standalone}
\begin{document}
\begin{tikzpicture}[even odd rule]
\draw (0,0) rectangle (2,2) (1.5,1.5) rectangle (.5,.5);
\begin{pgfinterruptboundingbox}
\clip (0,0) rectangle (2,2) (1.5,1.5) rectangle (.5,.5) (3.1,3.1) rectangle (-1.1,-1.1);
\end{pgfinterruptboundingbox}
\draw (-1,-1)--++(4,4);
\end{tikzpicture}
\end{document}

Output