smc

By Vieta's formulas, $c+d=-a$, $cd=b$, $a+b=-c$, and $ab=d$. From the equation $c+d=-a$, $d=-a-c$, and from the equation $a+b=-c$, $b=-a-c$, so $b=d$. Then from the equation $cd=b$, $cb=b$. Since $b$ is nonzero, we can divide both sides of the equation by $b$ to get $c=1$. Similarly, from the equation $ab=d$, $ab=b$, so $a=1$. Then $b=d=-a-c=-2$. Therefore, $a+b+c+d=1+(-2)+1+(-2)=\boxed{-2}$. The answer is (B).