A nontrivial everywhere continuous function with uncountably many roots? This is my first post on SE, forgive any blunders.
I am looking for an example of a function $f:\mathbb{R} \to \mathbb{R}$ which is continuous everywhere but has uncountably many roots ($x$ such that $f(x) = 0$). I am not looking for trivial examples such as $f = 0$ for all $x$. This is not a homework problem. I'd prefer a nudge in the right direction rather than an explicit example. 
Thanks!
Edit: Thanks all! I've constructed my example with your help.
 A: Even though you seem to be happy with the given answers, I can't resist pointing out the following construction which I already mentioned in this thread. If this example is already contained in one of the links provided in the other answer, I apologize for the duplication:
Choose a space-filling curve $c: \mathbb{R} \to \mathbb{R}^2$ and compose it with the projection $p$ to one of the coordinate axes. This gives you an example of a continuous and surjective function $f = p \circ c: \mathbb{R} \to \mathbb{R}$ all of whose pre-images are uncountable.
A: The roots of a continuous function is always a closed subset of $\mathbb{R}$ : $\{0\}$ is closed, thus $f^{-1}(\{0\})$ is closed too.
If you have a closed set $S$, you can define a function $f : x \mapsto d(x,S)$, which is continuous and whose set of roots is exactly $S$ : you can make a continuous function have any closed set as its set of roots.
Therefore you only have to look for closed sets that are uncountable.
A: This could be interesting for you.
A: If $f(x)$ is the function then $E = \{ x \colon f(x) = 0 \} $ is a closed set. Do you know of a nontrivial, by your standards, closed set with uncountable many points? The complement of $E$ is open. Is there a way to describe the complement of $E$ so that it would be easy to construct the remainder of the function in such a way that the function was never $0$ on the complement of $E$.
A: Why not $f(x)=x+|x|$? Looks quite nontrivial to me.
A: I just thought I'd mention that a Brownian motion has this property (almost surely).
A: If you just want a continuous function, then let $f(x) = 0$ over an interval say $[a,b]$ where $a<b$ and over the rest of the real line find functions $g(x)$ and $h(x)$ such that $g(a) = 0$ and $h(b) = 0$ and define $f(x) = g(x), \forall x \leq a$ and $f(x) = h(x), \forall x \geq b$.
You could also have infinitely smooth functions with zero on an uncountable set as for instance
$$f(x)=\begin{cases}e^{-1/x}&x>0\newline 0&x\leq 0\end{cases}$$
And if you want an infinitely smooth function with a compact support
$$
f(x) = 
\begin{cases}
\exp(- \frac{1}{1 - x^2}), \quad  |x| < 1 \\
0 , \quad |x| \geq 1

\end{cases}

$$
