[Tex/LaTex] TikZ block diagram with multiple arrows not from centre of block

Question

I'm trying to create a rather simple diagram but I don't know how to get the multiple arrows on the left not leaving the centre of the node.

Answer 1

Two options; using anchors and some shifts, and using the <name>.<angle > syntax:

\documentclass{article}
\usepackage{tikz}

\begin{document}

\begin{tikzpicture}
\node[draw,minimum size=2cm] (x) {X};
\draw[->] ([yshift=-10pt]x.west) -- node[fill=white] {a} +(-1cm,0pt);
\draw[->] ([yshift=10pt]x.west) -- node[fill=white] {b} +(-1cm,0pt);
\draw[->] (x.120) -- node[fill=white] {c} +(0pt,1cm);
\draw[->] (x.60) -- node[fill=white] {d} +(0pt,1cm);
\end{tikzpicture}

\end{document}

As Claudio Fiandrino has mentioned in his comment, another option is to use the calc library, so the shifts are not absolute, but can be calculated in terms of anchors:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}

\begin{tikzpicture}
\node[draw,minimum size=2cm] (x) {X};
\draw[->] ([yshift=-10pt]x.west) -- node[fill=white] {a} +(-1cm,0pt);
\draw[->] ([yshift=10pt]x.west) -- node[fill=white] {b} +(-1cm,0pt);
\draw[->] (x.120) -- node[fill=white] {c} +(0pt,1cm);
\draw[->] (x.60) -- node[fill=white] {d} +(0pt,1cm);
\draw[->]
  ( $ (x.north east)!0.5!(x.east) $ ) -- 
    node[fill=white] {e} 
  +(1cm,0pt);
\draw[->] 
  ( $ (x.east)!0.5!(x.south east) $ ) -- 
    node[fill=white] {f} 
    +(1cm,0pt);
\end{tikzpicture}

\end{document}

In the above example ( $ (x.north east)!0.5!(x.east) $ ) means the point whose coordinate is halfway between x.north east and x.east.