If there is a non-constructible real, is there an $L$-generic real? If we assume that $\Bbb R^V\neq\Bbb R^L$, can we deduce that there is some $x\in\Bbb R^V$ which is $L$-generic?
Of course if $V$ is a generic extension of $L$ this is true, but if $V=L[0^\#]$ this is also true (the existence of $0^\#$ implies the existence of $L$-Cohen generics), but $0^\#$ cannot be added by forcing. So it is possible to have reals which are not generic over $L$, but their existence does imply that a generic exist.
Is this always the case, or is there a counterexample? Namely, can we have $V=L[x]$ for some real number $x$ such that no real number in $L[x]$ is $L$-generic?
Edit: As Joel's answer shows, we can generate a counterexample using class forcing over $L$. To avoid these, we might as well require that the universe is not a class-generic extension of some $W$ for which $\Bbb R^W=\Bbb R^L$.
 A: Great question!
In order to formalize the notion, let us understand the phrase "$x$ is $L$-generic" to mean: there is some partial order $\mathbb{P}\in L$ and some filter $G\subset\mathbb{P}$ that is $L$-generic, such that $x\in L[G]$. In particular, this refers to set-sized forcing only.
In this case, we can give a negative answer. Sy Friedman has a way to undertake the coding-the-universe forcing in such a way that the generic extension $V[G]=L[R]$ is minimal: Minimal Coding, Annals of Pure and Applied Logic, 1989, pp. 233-297. In particular, every real in the extension is either in $V$, or generates $R$, which is not set-generic (although it is generic for class forcing). Thus, if you start in $L$ and then undertake this forcing, you get a class forcing extension in which there are no $L$-generic reals for set forcing, and indeed, no $L$-generic sets of any kind. 
In the linked paper, Friedman states the following:
Corollary. There is an $L$-definable forcing for producing a real $R$ which is minimal over $L$ but not set-generic over $L$. 
It follows that this extension $L[R]$ has no set-generic objects at all not in $L$. That is, it is a strong counterexample, which applies not just to reals, but to sets of any size.
