`Missing = inserted for \ifnum` when using addplot

Question

I'm trying to plot a normalized sinc function with PGFPlots/TikZ. As this is somewhat tricky because of the limiting value at 0, I used a suggestion provided by percusse.

Here's a minimal example (note that I've renamed the function to sincf due to conflicts with the rest of the document and inserted an approximation of Pi to normalize):

\documentclass{scrreprt}

\usepackage{pgfplots}
\pgfplotsset{compat=1.12}


\begin{document}
\begin{tikzpicture}

% https://tex.stackexchange.com/a/235009
\pgfmathdeclarefunction{sincf}{1}{%
        \pgfmathparse{abs(#1)<0.01 ? int(1) : int(0)}%
        \ifnum\pgfmathresult>0 %
            \pgfmathparse{1}%
        \else%
            \pgfmathparse{sin(3.14159*#1 r)/(3.14159*#1)}%
        \fi%
}

    \begin{axis}
        \addplot {sincf(\x)};
    \end{axis}

\end{tikzpicture}
\end{document}

However, I'm getting two errors which I can't figure out:

! Missing = inserted for \ifnum.
<to be read again>
Y
l.23 \addplot plot (\x,{sincf(\x)});
I was expecting to see `<', `=', or `>'. Didn't.

! Missing number, treated as zero.
<to be read again>
Y
l.23 \addplot plot (\x,{sincf(\x)});
A number should have been here; I inserted `0'.
(If you can't figure out why I needed to see a number,
look up `weird error' in the index to The TeXbook.)

Using MikTeX 2.9

Answer 1

pgfplots uses library fpu for better precision by default, which has a different internal number format. Therefore \pgfresult is not 1 or 0 anymore and \ifnum cannot be used.

\tikzset{fpu}
\pgfmathparse{int(0)}\show\pgfmathresult
\pgfmathparse{int(1)}\show\pgfmathresult

The lines show 0Y0 for zero and 1Y1 for one.

In this case, the whole math can be put in one \pgfmathparse expression:

\documentclass{scrreprt}

\usepackage{pgfplots}
\pgfplotsset{compat=1.12}

\begin{document}
\begin{tikzpicture}

    % http://tex.stackexchange.com/a/235009
    \pgfmathdeclarefunction{sincf}{1}{%
        \pgfmathparse{(abs(#1)<0.01) ? 1 : sin(pi*#1 r)/(pi*#1)}%
    }

    \begin{axis}
        \addplot {sincf(\x)};
    \end{axis}

\end{tikzpicture}
\end{document}

Answer 2

Unfortunately, both of the existing answers have drawbacks.

Heiko Oberdiek's answer: it exactly undoes the fix in the linked question. In this case \addplot just happens to not evaluate the function at x=0, but if it does...

\documentclass{article}
\usepackage{pgfplots}
\pgfplotsset{compat=1.17}
\begin{document}

    \pgfmathdeclarefunction{sincf}{1}{%
        \pgfmathparse{(abs(#1)<0.01) ? 1 : sin(pi*#1 r)/(pi*#1)}%
    }
    \pgfmathparse{sincf(0)}

\end{document}

you got an error.

Cryptc's answer works, but only if the FPU is currently enabled and the output format is float.

\documentclass{article}
\usepackage{pgfplots}
\pgfplotsset{compat=1.17}
\begin{document}

\pgfkeys{/pgf/fpu=true,/pgf/fpu/output format=fixed}
\pgfmathparse{1}  % stores 1.0000000000 to \pgfmathresult
\pgfmathfloattofixed{\pgfmathresult}  % \pgfmathfloattofixed, as with other \pgfmathfloat... functions, expect input in low-level PGF FPU representation

\end{document}

Error.

---

One way I can find to convert anything to an int is to round-trip through PGF's FPU:

\documentclass{article}
\usepackage{tikz}
% don't need fpu library!
\begin{document}

% works with both internal representation and "normal" representation

\pgfmathfloatparsenumber{1Y1.0e6]}
\pgfmathfloattoint\pgfmathresult
% now \pgfmathresult is always an integer

\pgfmathfloatparsenumber{1000000}
\pgfmathfloattoint\pgfmathresult

\end{document}

You can try \pgfmathtruncatemacro, but this will silently return the wrong error if the FPU is enabled and output format is float!

---

So for this problem, the fix is

\documentclass{scrreprt}

\usepackage{pgfplots}
\pgfplotsset{compat=1.12}

\begin{document}

\pgfmathdeclarefunction{sinc}{1}{%
    \pgfmathparse{abs(#1)<0.01 ? 1 : 0}%
    % ======== add these 2 lines ========
    \pgfmathfloatparsenumber\pgfmathresult
    \pgfmathfloattoint\pgfmathresult
    % ======== end ========
    \ifnum\pgfmathresult>0 %
        \pgfmathparse{1}%
    \else%
        \pgfmathparse{sin(3.14159*#1 r)/(3.14159*#1)}%
    \fi%
}

\begin{tikzpicture}
    \begin{axis}
        \addplot {sinc(\x)};
    \end{axis}
\end{tikzpicture}

% ======== double check it works with normal \draw plot
\begin{tikzpicture}
    \draw (-1,0) -- (1,0);
    \draw[domain=0:0.5, samples=250] plot (\x,{sinc(20*\x)});
\end{tikzpicture}

\end{document}

works both with TikZ \draw plot and \addplot.

---

Alternatively, staying completely within the FPU, there's \pgfmathfloatifflags:

\pgfmathdeclarefunction{sinc}{1}{%
    \pgfmathparse{abs(#1)<0.01}%
    \pgfmathfloatparsenumber\pgfmathresult
    \pgfmathfloatifflags{\pgfmathresult}{1}{% "less than" is true
        \pgfmathparse{1}%
    }{% "less than" is false
        \pgfmathparse{sin(3.14159*#1 r)/(3.14159*#1)}%
    }%
}

Also a bit faster.

---

Another alternative, although specific to this question, is

\pgfmathdeclarefunction{sinc}{1}{%
    \pgfmathparse{(#1==0 ? 1: sin(3.14159*#1 r))/(#1==0 ? 1: 3.14159*#1)}%
}

(just stare at the code for a while you'll see why it works.)

---

I tried to make ifthenelse short-circuits, but evaluate-immediately-while-parsing behavior is quite hard-wired into the behavior of \pgfmathparse, so it seems difficult to fix.

Answer 3

Using a related answer about fpu formatting, I was able to use \ifnum in the case that my result was within the range of an int() [−16383.9998, +16383.9998]. The trick is to use \pgfmathfloattofixed on the \pgfmathresult.

\pgfkeys{/pgf/fpu=true}
\pgfmathparse{...do floating point math here (with an int() result)...}
\pgfmathfloattofixed{\pgfmathresult} % this allows the conversion with int()
\edef\tmp{\pgfmathresult}
\pgfkeys{/pgf/fpu=false}
\pgfmathsetmacro\iTmp{int(\tmp)} % convert with int()

\ifnum \iTmp>0
  ... do stuff ...
\fi