# What exactly does the ! option do for floats

I finally found a case where adding ! to a float actually did something. Specifically, it allowed a fourth top float to appear on a page (with topnumber=3).

So, is that it?

\documentclass{article}
%\setcounter{topnumber}{4}

\makeatletter
\newcommand{\bitflags}{\def\showbitflag{\space\number\count}%
\let\@elt=\showbitflag
\@toplist}
\makeatother

\begin{document}
\begin{figure}[t]
\caption{created first}
\end{figure}

\begin{table}[t]
\caption{created second}
\end{table}

\begin{figure}[t!]
\caption{created third}
\end{figure}

\begin{table}[t!]
\caption{created fourth}
\end{table}

{\bitflags}
\end{document}


Normally a top [t] float is only added if some numeric constraints such as topnumber (the number of floats allowed at top of a page) are not violated. The ! tells latex to ignore these constraints (that is, for this one float act as if they had their most permissive values)