How to scroll over continuous functions? $\newcommand{\RR}{\mathbb{R}}$
$\newcommand{\CR}{\mathcal{C}\left(\RR\right)}$
There are as many continuous functions than real numbers: $\left|\left\{f, f:\CR \rightarrow \RR\right\}\right|=|\RR|$.
It is easy, for a machine, to approximately scroll over every real number from $\text{-}\infty$ to $\text{+}\infty$, provided we choose arbitrary, finite bounds and an arbitrary, finite precision. (btw cheers, Turing ;)
Since there are bijections between $\CR$ and $\RR$, is there not, similarly, a natural way to scroll over $\CR$? How would these —necessary— choices of "bounds and precision" translate?
My guess is that there may be many possible approaches (scroll over polynomials coefficients and degrees, over coefficients of periodic functions and the length of their sum, etc.), but then.. 
How come this problem seems more difficult than just $\textit{spanning}\ \RR$ since $\RR$ and $\CR$ are equipotent?
How come it seems to need more variables and more arbitrary choices?

[EDIT:] as suggested in the comments, here would be the requirements for such a "natural scrolling":
Let $s:\left\{\begin{array}{ll}
\RR \to \CR \\
r \mapsto s_r
\end{array}\right.$ be called a "scrolling" of $\CR$.
$\forall x \in \RR$, we can define a trace function, tracking the evolution of the image of $x$ by $s_r$ while scrolling: $t_x:\left\{\begin{array}{ll}
\RR \to \RR \\
r \mapsto s_r(x)
\end{array}\right.$
Can we have both:


*

*$s\ \text{bijective}$

*$t_x \in \CR \forall x \in \RR$ ?


Just an observation. 
Even tough all trace maps are continuous, the diagonalized trace map 
 $x\mapsto t_x(x)$ cannot be continuous. If it were continuous, then so would $f(x)=t_x(x)+1$. So there would be some $r\in\Bbb R$ with $s(r)=f$. But then
$$f(r)=t_r(r)+1=\phi(r)(r)+1=f(r)+1,$$
Contradiction.
