1. For all real numbers $x, y$, if the function $f$ satisfies:
$$
f(x y)=f(x) \cdot f(y)
$$

and $f(0) \neq 0$, then $f(1998)=$ $\qquad$ .
$\approx 、 1.1$.
Let $f(x y)=f(x) \cdot f(y)$, set $y=0$, we have $f(x \cdot 0)=$ $f(x) \cdot f(0)$, which means $f(0)=f(x) \cdot f(0)$, and since $f(0) \neq 0$, it follows that $f(x)=1$, thus $f(1998)=1$.