38    world is not ideal), and one cycle per instruction, then 43+3*(n/48-1)
39    <= 24+24*(n/48-1) so n >= 45.7; n >= 0.9; we win on the first full
55   register int n __asm__ ("r12") = plen;  in memset()
63   __asm__("movu.b %0,r13						\n\  in memset()
64 	   lslq 8,r13							\n\  in memset()
65 	   move.b %0,r13						\n\  in memset()
66 	   move.d r13,%0						\n\  in memset()
67 	   lslq 16,r13							\n\  in memset()
77 	/* Oops! n = 0 must be a valid call, regardless of alignment.  */  in memset()
78 	&& n >= 3)  in memset()
83 	    n--;  in memset()
90 	    n -= 2;  in memset()
96     if (n >= MEMSET_BY_BLOCK_THRESHOLD)  in memset()
104 	   ;; GCC does promise correct register allocations, but let's	\n\  in memset()
105 	   ;; make sure it keeps its promises.				\n\  in memset()
106 	   .ifnc %0-%1-%4,$r13-$r12-$r11				\n\  in memset()
107 	   .error \"GCC reg alloc bug: %0-%1-%4 != $r13-$r12-$r11\"	\n\  in memset()
108 	   .endif							\n\  in memset()
109 									\n\  in memset()
110 	   ;; Save the registers we'll clobber in the movem process	\n\  in memset()
111 	   ;; on the stack.  Don't mention them to gcc, it will only be	\n\  in memset()
112 	   ;; upset.							\n\  in memset()
113 	   subq	   11*4,sp						\n\  in memset()
114 	   movem   r10,[sp]						\n\  in memset()
115 									\n\  in memset()
116 	   move.d  r11,r0						\n\  in memset()
117 	   move.d  r11,r1						\n\  in memset()
118 	   move.d  r11,r2						\n\  in memset()
119 	   move.d  r11,r3						\n\  in memset()
120 	   move.d  r11,r4						\n\  in memset()
121 	   move.d  r11,r5						\n\  in memset()
122 	   move.d  r11,r6						\n\  in memset()
123 	   move.d  r11,r7						\n\  in memset()
124 	   move.d  r11,r8						\n\  in memset()
125 	   move.d  r11,r9						\n\  in memset()
126 	   move.d  r11,r10						\n\  in memset()
127 									\n\  in memset()
128 	   ;; Now we've got this:					\n\  in memset()
129 	   ;; r13 - dst							\n\  in memset()
130 	   ;; r12 - n							\n\  in memset()
131 									\n\  in memset()
132 	   ;; Update n for the first loop				\n\  in memset()
133 	   subq	   12*4,r12						\n\  in memset()
134 0:									\n\  in memset()
141 "	   setf\n"  in memset()
143 "	   subq	  12*4,r12						\n\  in memset()
144 	   bge	   0b							\n\  in memset()
145 	   movem	r11,[r13+]					\n\  in memset()
146 									\n\  in memset()
147 	   ;; Compensate for last loop underflowing n.			\n\  in memset()
148 	   addq	  12*4,r12						\n\  in memset()
149 									\n\  in memset()
150 	   ;; Restore registers from stack.				\n\  in memset()
154 	   : "=r" (dst), "=r" (n)  in memset()
157 	   : "0" (dst), "1" (n), "r" (lc));  in memset()
161     while (n >= 16)  in memset()
167 	n -= 16;  in memset()
170     switch (n)  in memset()