ve example parser <tt>3 + 4 * 8</tt>, the parser doesn't know
whether to reduce <tt>3 + 4</tt> or shift the '*' onto the stack.
However, since '*' has a higher precedence than '+', it will be shifted
and the multiplication will be performed before the addition.<p>

CUP assigns each one of its terminals a precedence according to these
declarations.  Any terminals not in this declaration have lowest
precedence.  CUP also assigns each of its productions a precedence.
That precedence is equal to the precedence 