How to declare "newcommand" value as a sum of some number (with units) and another command so that it can be used in coordinate arithmetic?

Question

I suspect that arithmetic I put inside newcommand is not processed as I would expect.

What is the value of \FMARGIN after declaring it as

\newcommand{\FWIDTH}{2mm}
\newcommand{\FMARGIN}{10mm+\FWIDTH}

Is it 12mm or something else?

Because when I use it inside coordinate arithmetic I get unexpected results. Not all margins are same distance away from the bounding box.

Complete code:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary
    {%
        calc,
        shapes.geometric
    }
\newcommand{\FWIDTH}{2mm}
\newcommand{\FMARGIN}{10mm+\FWIDTH}
\begin{document}
    \begin{tikzpicture}
        \path node
            [%
                isosceles triangle,
                draw=blue,
                minimum width=30mm,
                line width=1mm
            ]   {};
        \path
            [%
                draw=red,
                line width=1mm
            ]
            ($(current bounding box.north west)+(-\FMARGIN,\FMARGIN)$)
            rectangle
            ($(current bounding box.south east)+(\FMARGIN,-\FMARGIN)$);
    \end{tikzpicture}
\end{document}

Output:

Difference becomes more obvious when \FWIDTH is set to a higher value (like 5mm).

Answer 1

What was originally used in the question, namely

\newcommand{\FMARGIN}{10mm+\FWIDTH}

stores as \FMARGIN the tokens in the command. Substitution and evaluation is not made until it is used. However, when TeX is looking for a length, something like 10mm+2mm will not do. Rather, the command \dimexpr10mm+2mm\relax is needed to perform the addition of lengths and provide the result as a single length which can be used by the macro seeking a length input.

\dimexpr stands for "dimensional expression" (the dimension being length) and \relax terminates the evaluation of a dimensional expression.

If one needs the result in textual form, that is, the tokens that are equivalent to a length of 12mm, then one uses \the\dimexpr10mm+2mm\relax, whereas without the \the, the result is stored as an internal length and not a series of alphanumeric tokens.

A good way to see what is happening is comparing this code:

\edef\tmp{10mm+\FWIDTH}
\detokenize\expandafter{\tmp}

to this code

\edef\tmp{\the\dimexpr10mm+\FWIDTH\relax}
\detokenize\expandafter{\tmp}

The former evaluates to 10mm+2mm while the latter evaluates to 34.14328pt. Here is the MWE.

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary
    {%
        calc,
        shapes.geometric
    }
\newcommand{\FWIDTH}{2mm}
\newcommand{\FMARGIN}{\dimexpr10mm+\FWIDTH\relax}
\begin{document}
    \begin{tikzpicture}
        \path node
            [%
                isosceles triangle,
                draw=blue,
                minimum width=30mm,
                line width=1mm
            ]   {};
        \path
            [%
                draw=red,
                line width=1mm
            ]
            ($(current bounding box.north west)+(-\FMARGIN,\FMARGIN)$)
            rectangle
            ($(current bounding box.south east)+(\FMARGIN,-\FMARGIN)$);
    \end{tikzpicture}
\end{document}

Answer 2

When defining macros (such as with \newcommand), TeX does no arithmetic: you can think of it as simple substitution of strings (or to be more precise, tokens). So when you write:

\newcommand{\FWIDTH}{2mm}
\newcommand{\FMARGIN}{10mm+\FWIDTH}

no arithmetic happens in 10mm+\FWIDTH. The result is that \FMARGIN gets defined as a macro which expands to 10mm+\FWIDTH (and this has no special arithmetical meaning; it's simply a sequence of five tokens 1, 0, m, m, +, followed by the token \FWIDTH). And when you use it somewhere as \FMARGIN, under typical circumstances it will (ultimately) expand to the string 10mm+2mm, which again is simply a sequence of 8 characters: it's as if you had typed 10mm+2mm at that place.

So when you use:

        ($(current bounding box.north west)+(-\FMARGIN,\FMARGIN)$)
        rectangle
        ($(current bounding box.south east)+(\FMARGIN,-\FMARGIN)$);

it's as if you had typed

        ($(current bounding box.north west)+(-10mm+2mm,10mm+2mm)$)
        rectangle
        ($(current bounding box.south east)+(10mm+2mm,-10mm+2mm)$);

and although this happens to compile, it is clearly not what you want. Macros work this way because TeX was initially designed for typesetting, and macros were supposed to be a shortcut for saving typing, not for doing computation.

You can debug this for yourself (see what has \FMARGIN been defined as) by adding \show\FMARGIN) into the file:

\newcommand{\FWIDTH}{2mm}
\newcommand{\FMARGIN}{10mm+\FWIDTH}
\show\FMARGIN

when TeX reaches that point, it will show you the expansion and wait for you to hit Return before continuing:

> \FMARGIN=\long macro:
->10mm+\FWIDTH .
l.10 \show\FMARGIN

? 

Reading this requires some training, but what the above tells you is that \FMARGIN is defined as 10mm+\FWIDTH (not as 12mm).

So how do you fix this? In this case, because mm is a dimension that TeX happens to understand (but only when it's looking for a length, not when simply defining macros and such), there are some tricks you can use:

• you can define \FMARGIN as a length, using \setlength, as in @cfr's answer

• you can define \FMARGIN to expand to something that will result in 12mm when TeX is looking for a length, as in @Steven's answer

• you can implement your original intention of somehow defining \FMARGIN to expand to 12mm, though it is slightly tricky (again, see Steven's answer).

Answer 3

This isn't an answer, but too long for a comment. It doesn't answer the question asked, that is. It does address a potential follow-up of how to avoid the problem.

Macros expand to their replacement text. For handling dimensions, it is easiest to use dimensions/lengths. For handling integers, it is easiest to use counts/counters.

\documentclass{standalone}
\usepackage{tikz,calc}
\usetikzlibrary{calc,shapes.geometric}
\newlength\FMARGIN
\newlength\FWIDTH
\tikzset{%
  fwidth/.code={\setlength\FWIDTH{#1}},
  fmargin/.code={\setlength\FMARGIN{10mm+#1}},
  fwidth/.forward to=/tikz/fmargin,
  fwidth=2mm,
}
\begin{document}
\begin{tikzpicture}
  \path node [isosceles triangle, draw=blue, minimum width=30mm, line width=1mm ]   {};
  \path [draw=red, line width=1mm ] ($(current bounding box.north west)+(-\FMARGIN,\FMARGIN)$) rectangle ($(current bounding box.south east)+(\FMARGIN,-\FMARGIN)$);
\end{tikzpicture}
\begin{tikzpicture}[fwidth=20mm]
  \path node [isosceles triangle, draw=blue, minimum width=30mm, line width=1mm ]   {};
  \path [draw=red, line width=1mm ] ($(current bounding box.north west)+(-\FMARGIN,\FMARGIN)$) rectangle ($(current bounding box.south east)+(\FMARGIN,-\FMARGIN)$);
\end{tikzpicture}
\end{document}

Answer 4

I found \pgfmathsetlengthmacro to be the easiest solution.

All I did was replace \newcommand in \newcommand{\FWIDTH}{2mm} \newcommand{\FMARGIN}{10mm+\FWIDTH}

with \pgfmathsetlengthmacro to get \pgfmathsetlengthmacro{\FWIDTH}{2mm} \pgfmathsetlengthmacro{\FMARGIN}{10mm+\FWIDTH}

Now coordinate arithmetic calculates \FMARGIN correctly.