zlib/intel/adler32vec.s

7 	Given 2 unsigned 32-bit alder and sum2 (both pre-modulo by BASE=65521) and a sequence of input bytes x[0],...x[N-1].
11 			adler = (adler+x[i])%BASE;
12 			sum2 = (sum2+adler)%BASE;
15 	To reduce/save the modulo operations, it can be shown that, if initial alder and sum2 are less than BASE(=65521),
24 			adler%=BASE;
25 			sum2%=BASE;
74 #define BASE 65521  /* largest prime smaller than 65536 */
75 #define NMAX 5552 	/* NMAX is the largest n such that 255n(n+1)/2 + (n+1)(BASE-1) <= 2^32-1 */
148 	movl		$$-2146992015, %eax		// 1/BASE in Q47
149 	mull		adler					// edx:eax = adler divided by BASE in Q47
150 	shrl		$$15, %edx				// edx is now the floor integer of adler and BASE
151 	imull		$$BASE, %edx, %edx		// edx * BASE
152 	subl		%edx, adler				// adler -= edx*BASE
153 	movl		$$-2146992015, %eax		// 1/BASE in Q47
154 	mull		sum2					// edx:eax = sum2 divided by BASE in Q47
155 	shrl		$$15, %edx				// edx is now the floor integer of sum2 and BASE
156 	imull		$$BASE, %edx, %eax		// eax = edx * BASE
157 	subl		%eax, sum2				// sum2 -= sdx*BASE
280 	modulo_BASE						// 		(adler/sum2) modulo BASE;
314 	modulo_BASE						// (adler/sum2) modulo BASE;
374 	modulo_BASE						// 		(adler/sum2) modulo BASE;
509 	movl		$$-2146992015, %eax		// 1/BASE in Q47
510 	mull		%edi					// edx:eax = adler divided by BASE in Q47
511 	shrl		$$15, %edx				// edx is now the floor integer of adler and BASE
512 	imull		$$BASE, %edx, %edx		// edx * BASE
513 	subq		%rdx, adler				// adler -= edx*BASE
514 	movl		$$-2146992015, %eax		// 1/BASE in Q47
515 	mull		%esi					// edx:eax = sum2 divided by BASE in Q47
516 	shrl		$$15, %edx				// edx is now the floor integer of sum2 and BASE
517 	imull		$$BASE, %edx, %eax		// eax = edx * BASE
518 	subq		%rax, sum2				// sum2 -= sdx*BASE
577 	modulo_BASE						// 		(adler/sum2) modulo BASE;
603 	modulo_BASE						// (adler/sum2) modulo BASE;
896 	modulo_BASE						// 		(adler/sum2) modulo BASE;
937 	modulo_BASE						// (adler/sum2) modulo BASE;