Function $f:\mathbb{R}\mapsto \mathbb{R}$ satisfying $f(f(x))=x+\lfloor x\rfloor$

Question

I'm looking for a function $f$ from reals to reals such that $$f(f(x))=x+\lfloor x\rfloor$$ ...or, in other words, a functional square root of the function $x+\lfloor x\rfloor$. I've been able to sketch out some parts of the graph of a function solving this equation, but I'm looking for a nice closed form, not a graph - especially one using the floor function.

NEAR MISS: The function $$g_{a}(x)=(a^2+ 2a)\lfloor x\rfloor +x$$ has functional square root $f_{a}(x)=a\lfloor x\rfloor+ x$, but only as long as $a$ is an integer, so I can't use it on the function I'm asking for.

Answer 1

A somewhat different answer to the amazing @RobertIsrael one: $$f(x)= \cases{x+\frac12 & if $x-\lfloor x\rfloor<\frac12 $\cr x+\lfloor x\rfloor-\frac12 & otherwise}$$

Answer 2

First you want to define $f:\;\mathbb Z \to \mathbb Z$ so that $f(f(n)) = 2n$, then for $x \in \mathbb R$ you can take $f(x) = f(\lfloor x \rfloor) + x - \lfloor x \rfloor$.

Take $f(0)=0$. For $n = 3^j k$ where $k$ is an integer coprime to $3$, let $$f(n) = \cases{3 n & if $j$ is even\cr 2 n/3 & if $j$ is odd}$$