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📄 div64.c

📁 linux 内核源代码
💻 C
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/* *  arch/s390/lib/div64.c * *  __div64_32 implementation for 31 bit. * *    Copyright (C) IBM Corp. 2006 *    Author(s): Martin Schwidefsky (schwidefsky@de.ibm.com), */#include <linux/types.h>#include <linux/module.h>#ifdef CONFIG_MARCH_G5/* * Function to divide an unsigned 64 bit integer by an unsigned * 31 bit integer using signed 64/32 bit division. */static uint32_t __div64_31(uint64_t *n, uint32_t base){	register uint32_t reg2 asm("2");	register uint32_t reg3 asm("3");	uint32_t *words = (uint32_t *) n;	uint32_t tmp;	/* Special case base==1, remainder = 0, quotient = n */	if (base == 1)		return 0;	/*	 * Special case base==0 will cause a fixed point divide exception	 * on the dr instruction and may not happen anyway. For the	 * following calculation we can assume base > 1. The first	 * signed 64 / 32 bit division with an upper half of 0 will	 * give the correct upper half of the 64 bit quotient.	 */	reg2 = 0UL;	reg3 = words[0];	asm volatile(		"	dr	%0,%2\n"		: "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" );	words[0] = reg3;	reg3 = words[1];	/*	 * To get the lower half of the 64 bit quotient and the 32 bit	 * remainder we have to use a little trick. Since we only have	 * a signed division the quotient can get too big. To avoid this	 * the 64 bit dividend is halved, then the signed division will	 * work. Afterwards the quotient and the remainder are doubled.	 * If the last bit of the dividend has been one the remainder	 * is increased by one then checked against the base. If the	 * remainder has overflown subtract base and increase the	 * quotient. Simple, no ?	 */	asm volatile(		"	nr	%2,%1\n"		"	srdl	%0,1\n"		"	dr	%0,%3\n"		"	alr	%0,%0\n"		"	alr	%1,%1\n"		"	alr	%0,%2\n"		"	clr	%0,%3\n"		"	jl	0f\n"		"	slr	%0,%3\n"		"	alr	%1,%2\n"		"0:\n"		: "+d" (reg2), "+d" (reg3), "=d" (tmp)		: "d" (base), "2" (1UL) : "cc" );	words[1] = reg3;	return reg2;}/* * Function to divide an unsigned 64 bit integer by an unsigned * 32 bit integer using the unsigned 64/31 bit division. */uint32_t __div64_32(uint64_t *n, uint32_t base){	uint32_t r;	/*	 * If the most significant bit of base is set, divide n by	 * (base/2). That allows to use 64/31 bit division and gives a	 * good approximation of the result: n = (base/2)*q + r. The	 * result needs to be corrected with two simple transformations.	 * If base is already < 2^31-1 __div64_31 can be used directly.	 */	r = __div64_31(n, ((signed) base < 0) ? (base/2) : base);	if ((signed) base < 0) {		uint64_t q = *n;		/*		 * First transformation:		 * n = (base/2)*q + r		 *   = ((base/2)*2)*(q/2) + ((q&1) ? (base/2) : 0) + r		 * Since r < (base/2), r + (base/2) < base.		 * With q1 = (q/2) and r1 = r + ((q&1) ? (base/2) : 0)		 * n = ((base/2)*2)*q1 + r1 with r1 < base.		 */		if (q & 1)			r += base/2;		q >>= 1;		/*		 * Second transformation. ((base/2)*2) could have lost the		 * last bit.		 * n = ((base/2)*2)*q1 + r1		 *   = base*q1 - ((base&1) ? q1 : 0) + r1		 */		if (base & 1) {			int64_t rx = r - q;			/*			 * base is >= 2^31. The worst case for the while			 * loop is n=2^64-1 base=2^31+1. That gives a			 * maximum for q=(2^64-1)/2^31 = 0x1ffffffff. Since			 * base >= 2^31 the loop is finished after a maximum			 * of three iterations.			 */			while (rx < 0) {				rx += base;				q--;			}			r = rx;		}		*n = q;	}	return r;}#else /* MARCH_G5 */uint32_t __div64_32(uint64_t *n, uint32_t base){	register uint32_t reg2 asm("2");	register uint32_t reg3 asm("3");	uint32_t *words = (uint32_t *) n;	reg2 = 0UL;	reg3 = words[0];	asm volatile(		"	dlr	%0,%2\n"		: "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" );	words[0] = reg3;	reg3 = words[1];	asm volatile(		"	dlr	%0,%2\n"		: "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" );	words[1] = reg3;	return reg2;}#endif /* MARCH_G5 */

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