Direct evaluation of fp expression

Question

This code

\documentclass{article}
\usepackage[nomessages]{fp}
\begin{document}
\FPeval{result}{clip(cos(pi))}
$\cos(\pi)=$\result
\end{document}

would produce the following output:

Is it possible to use the result of fp expressions directly without the necessity to have them pre-computed and stored in a permanent variable so that I could transform the code to the following form:

\documentclass{article}
\usepackage[nomessages]{fp}
\begin{document}
$\cos(\pi)=$<fp expression>
\end{document}

Answer 1

Not sure if this approach has any gotcha's.

\documentclass{article}
\usepackage[nomessages]{fp}
\newcommand\FPuse[1]{\FPeval{\result}{#1}{\result}}
\begin{document}
$\cos(\pi)=\FPuse{clip(cos(pi))}$\

$\sin(\pi/3)=\FPuse{sin(pi/3)}$

$\sin(\pi/3)=\FPuse{round(sin(pi/3),3)}$
\end{document}

In the comments below, jfbu and I discuss why I grouped {\result} at the end of the \FPuse definition. First, let's see what happens if I ungroup it:

\newcommand\FPuse[1]{\FPeval{\result}{#1}\result}

The result on the first operation is

What we see is that \result sets itself as {} - 1, using a binary minus sign. The conclusion is that \FPeval{}{} creates a \bgroup...\egroup quantity that, in math mode, causes the subsequent minus sign to act in a binary fashion. Thus, the only way to eliminate this problem (without changing the fp package), is to isolate the final \result in its own group, as I did in my original code.

While jfbu has probed a bit into the guts of fp, I am no expert to know if the fp code can be revised to use \begingroup...\endgroup instead (which is truly transparent in math mode) or not. I do know that fp has a few issues, for example, a stray space is introduced via \FPpow which has to be \unskiped after its use.

See jfbu's comments below for more information on the group issue.

Answer 2

Not with fp. It is possible with expl3, though.

\documentclass{article}
\usepackage{expl3}

\ExplSyntaxOn
% make an internal function available to the user
\cs_set_eq:NN \fpeval \fp_eval:n
\ExplSyntaxOff

\begin{document}

$\cos(\pi)=\fpeval{cos(pi)}$

$\sin(\pi/3)=\fpeval{sin(pi/3)}$

$\sin(\pi/3)=\fpeval{round(sin(pi/3),3)}$

\end{document}