50  *	v = 2^n v1  *  v0
54  *	uv = 2^2n u1 v1  +  2^n u1 v0  +  2^n v1 u0  +  u0 v0
55  *	   = 2^2n u1 v1  +     2^n (u1 v0 + v1 u0)   +  u0 v0
57  * Now add 2^n u1 v1 to the first term and subtract it from the middle,
61  *	uv = (2^2n + 2^n) (u1 v1)  +
62  *	         (2^n)    (u1 v0 - u1 v1 + u0 v1 - u0 v0)  +
67  *	uv = (2^2n + 2^n) (u1 v1)  +			[u1v1 = high]
68  *		 (2^n)    (u1 - u0) (v0 - v1)  +	[(u1-u0)... = mid]
71  * The terms (u1 v1), (u1 - u0) (v0 - v1), and (u0 v0) can all be done
73  * of (u1 - u0) or (v0 - v1) may be negative.)
108 #define	v1	v.ul[H]  in __muldi3()  macro
113 	 * u1, u0, v1, and v0 will be directly accessible through the  in __muldi3()
125 	if (u1 == 0 && v1 == 0) {  in __muldi3()
127 		 * An (I hope) important optimization occurs when u1 and v1  in __muldi3()
145 		if (v0 >= v1)  in __muldi3()
146 			vdiff = v0 - v1;  in __muldi3()
148 			vdiff = v1 - v0, negmid ^= 1;  in __muldi3()
151 		high = u1 * v1;  in __muldi3()
163 #undef v1  in __muldi3()
187 	u_int u1, u0, v1, v0, udiff, vdiff, high, mid, low;  in __lmulq()  local
194 	v1 = HHALF(v);  in __lmulq()
200 	if (u1 == 0 && v1 == 0)  in __lmulq()
207 	if (v0 >= v1)  in __lmulq()
208 		vdiff = v0 - v1;  in __lmulq()
210 		vdiff = v1 - v0, neg ^= 1;  in __lmulq()
213 	high = u1 * v1;  in __lmulq()