[Tex/LaTex] How to increment variable by using same variable

counterspgfmath

How can I increment variable from the same variable?

\pgfmathsetmacro\S{5};
\pgfmathsetmacro\S{\S + 1};%   not working

How can I work around this? I need counters which I use as line coordinates increment in certain conditions.

Update

\pgfmathsetmacro\cA{0}; 
\newcounter{cB}
\foreach \x in {1,...,10}
{ 
    \pgfmathtruncatemacro\cA{\cA+1)};
    \pgfmathaddtocounter{cB}{1};            
    \node at (\x,1) { \cA };
    \node at (\x,0) { \the\numexpr\value{cB} };         
}

printed out this

1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1

I need

1 2 3 4 ...

Yes, I could do this in this simple example by just using \x variable but in my real diagram I need to increment them irregularly. So I need variable which can be incremented inside the loop without resets. Or, am I missing something and it should work?

Best Answer

To be used in a \foreach loop, there are better options:

\documentclass[tikz,border=2pt]{standalone}
\begin{document}

\begin{tikzpicture}
\foreach \i [count=\S from 5] in {1,...,5}
    \node [draw, xshift=\i cm] {\S};
\end{tikzpicture}

\end{document}

where the syntax count=\S from 5 is used here to set \S to 5 and advance it by 1 in each iteration. Another syntax may be evaluate=\i as \S using \i+4, which will achieve the same result.

Update

The increment can be changed within the loop based on a condition like this:

\newcounter{cA} 
\setcounter{cA}{0}
\newcounter{cB}
\setcounter{cB}{0}

\begin{tikzpicture}
\foreach \x in {1,...,10}{ 
    \addtocounter{cA}{1}
    \ifnum\x<6\addtocounter{cB}{1}\else\addtocounter{cB}{2}\fi            
    \node at (\x,1) { \thecA };
    \node at (\x,0) { \thecB };          
}
\end{tikzpicture}

Related Solutions

[Tex/LaTex] Increment loop variable in inner foreach loop

You're right, it's a bit of a shame that the expressions aren't evaluated before the loop is started.

Three approaches:

Use the count=\xi expression to make the outer loop counter accessible. In this case, the outer loop only has to run from 2 to 4, while the counter will run from 1 to 3, which happens to be the starting point of the inner loop. This solution only works in this special case, though, as soon as you need the inner loop to be larger by a different value, you'll have to resort to something else.

Alternatively, you can parse the expression at the start of the outer loop using \pgfmathtruncatemacro, which will save the result as an integer, or using the evaluate=\x as \result using <expression> syntax.

Note that the outer loop actually only needs to run from 1 to 3.

\documentclass{article}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\foreach \x [count=\xi] in {2,...,4} {
  \foreach \y  in {\x,...,4} {
    \node at (\xi,\y) {(\xi,\y)};
  }
}
\end{tikzpicture}

\begin{tikzpicture}
\foreach \x in {1,...,3} {
    \pgfmathtruncatemacro\ystart{\x+1}
    \foreach \y in {\ystart,...,4} {
      \node at (\x,\y) {(\x,\y)};
    }
}
\end{tikzpicture}

\begin{tikzpicture}
\foreach \x [evaluate=\x as \ystart using int(\x+1)] in {1,...,3} {
    \foreach \y in {\ystart,...,4} {
      \node at (\x,\y) {(\x,\y)};
    }
}
\end{tikzpicture}
\end{document}

Best Answer

Related Solutions

[Tex/LaTex] Increment loop variable in inner foreach loop

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