6. Let the function $y=f(x)$ satisfy $f(x+1)=f(x)+1$. Then the number of roots of the equation $f(x)=x$ could be ( ).
(A) infinitely many
(B) none or a finite number
(C) a finite number
(D) none or infinitely many
6.D.
$f(x+1)=f(x)+1$ obviously has a solution $f(x)=x+c$, where $c$ is any real number.
When $c \neq 0$, $f(x)=x$ has no solution;
When $c=0$, $f(x)=x$ has infinitely many solutions.