| Martin Schwidefsky | d9f7a74 | 2006-09-28 16:55:39 +0200 | [diff] [blame] | 1 | /* | 
 | 2 |  *  arch/s390/lib/div64.c | 
 | 3 |  * | 
 | 4 |  *  __div64_32 implementation for 31 bit. | 
 | 5 |  * | 
 | 6 |  *    Copyright (C) IBM Corp. 2006 | 
 | 7 |  *    Author(s): Martin Schwidefsky (schwidefsky@de.ibm.com), | 
 | 8 |  */ | 
 | 9 |  | 
 | 10 | #include <linux/types.h> | 
 | 11 | #include <linux/module.h> | 
 | 12 |  | 
 | 13 | #ifdef CONFIG_MARCH_G5 | 
 | 14 |  | 
 | 15 | /* | 
 | 16 |  * Function to divide an unsigned 64 bit integer by an unsigned | 
 | 17 |  * 31 bit integer using signed 64/32 bit division. | 
 | 18 |  */ | 
 | 19 | static uint32_t __div64_31(uint64_t *n, uint32_t base) | 
 | 20 | { | 
 | 21 | 	register uint32_t reg2 asm("2"); | 
 | 22 | 	register uint32_t reg3 asm("3"); | 
 | 23 | 	uint32_t *words = (uint32_t *) n; | 
 | 24 | 	uint32_t tmp; | 
 | 25 |  | 
 | 26 | 	/* Special case base==1, remainder = 0, quotient = n */ | 
 | 27 | 	if (base == 1) | 
 | 28 | 		return 0; | 
 | 29 | 	/* | 
 | 30 | 	 * Special case base==0 will cause a fixed point divide exception | 
 | 31 | 	 * on the dr instruction and may not happen anyway. For the | 
 | 32 | 	 * following calculation we can assume base > 1. The first | 
 | 33 | 	 * signed 64 / 32 bit division with an upper half of 0 will | 
 | 34 | 	 * give the correct upper half of the 64 bit quotient. | 
 | 35 | 	 */ | 
 | 36 | 	reg2 = 0UL; | 
 | 37 | 	reg3 = words[0]; | 
 | 38 | 	asm volatile( | 
 | 39 | 		"	dr	%0,%2\n" | 
 | 40 | 		: "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" ); | 
 | 41 | 	words[0] = reg3; | 
 | 42 | 	reg3 = words[1]; | 
 | 43 | 	/* | 
 | 44 | 	 * To get the lower half of the 64 bit quotient and the 32 bit | 
 | 45 | 	 * remainder we have to use a little trick. Since we only have | 
 | 46 | 	 * a signed division the quotient can get too big. To avoid this | 
 | 47 | 	 * the 64 bit dividend is halved, then the signed division will | 
 | 48 | 	 * work. Afterwards the quotient and the remainder are doubled. | 
 | 49 | 	 * If the last bit of the dividend has been one the remainder | 
 | 50 | 	 * is increased by one then checked against the base. If the | 
 | 51 | 	 * remainder has overflown subtract base and increase the | 
 | 52 | 	 * quotient. Simple, no ? | 
 | 53 | 	 */ | 
 | 54 | 	asm volatile( | 
 | 55 | 		"	nr	%2,%1\n" | 
 | 56 | 		"	srdl	%0,1\n" | 
 | 57 | 		"	dr	%0,%3\n" | 
 | 58 | 		"	alr	%0,%0\n" | 
 | 59 | 		"	alr	%1,%1\n" | 
 | 60 | 		"	alr	%0,%2\n" | 
 | 61 | 		"	clr	%0,%3\n" | 
 | 62 | 		"	jl	0f\n" | 
 | 63 | 		"	slr	%0,%3\n" | 
 | 64 | 		"	alr	%1,%2\n" | 
 | 65 | 		"0:\n" | 
 | 66 | 		: "+d" (reg2), "+d" (reg3), "=d" (tmp) | 
 | 67 | 		: "d" (base), "2" (1UL) : "cc" ); | 
 | 68 | 	words[1] = reg3; | 
 | 69 | 	return reg2; | 
 | 70 | } | 
 | 71 |  | 
 | 72 | /* | 
 | 73 |  * Function to divide an unsigned 64 bit integer by an unsigned | 
 | 74 |  * 32 bit integer using the unsigned 64/31 bit division. | 
 | 75 |  */ | 
 | 76 | uint32_t __div64_32(uint64_t *n, uint32_t base) | 
 | 77 | { | 
 | 78 | 	uint32_t r; | 
 | 79 |  | 
 | 80 | 	/* | 
 | 81 | 	 * If the most significant bit of base is set, divide n by | 
 | 82 | 	 * (base/2). That allows to use 64/31 bit division and gives a | 
 | 83 | 	 * good approximation of the result: n = (base/2)*q + r. The | 
 | 84 | 	 * result needs to be corrected with two simple transformations. | 
 | 85 | 	 * If base is already < 2^31-1 __div64_31 can be used directly. | 
 | 86 | 	 */ | 
 | 87 | 	r = __div64_31(n, ((signed) base < 0) ? (base/2) : base); | 
 | 88 | 	if ((signed) base < 0) { | 
 | 89 | 		uint64_t q = *n; | 
 | 90 | 		/* | 
 | 91 | 		 * First transformation: | 
 | 92 | 		 * n = (base/2)*q + r | 
 | 93 | 		 *   = ((base/2)*2)*(q/2) + ((q&1) ? (base/2) : 0) + r | 
 | 94 | 		 * Since r < (base/2), r + (base/2) < base. | 
 | 95 | 		 * With q1 = (q/2) and r1 = r + ((q&1) ? (base/2) : 0) | 
 | 96 | 		 * n = ((base/2)*2)*q1 + r1 with r1 < base. | 
 | 97 | 		 */ | 
 | 98 | 		if (q & 1) | 
 | 99 | 			r += base/2; | 
 | 100 | 		q >>= 1; | 
 | 101 | 		/* | 
 | 102 | 		 * Second transformation. ((base/2)*2) could have lost the | 
 | 103 | 		 * last bit. | 
 | 104 | 		 * n = ((base/2)*2)*q1 + r1 | 
 | 105 | 		 *   = base*q1 - ((base&1) ? q1 : 0) + r1 | 
 | 106 | 		 */ | 
 | 107 | 		if (base & 1) { | 
 | 108 | 			int64_t rx = r - q; | 
 | 109 | 			/* | 
 | 110 | 			 * base is >= 2^31. The worst case for the while | 
 | 111 | 			 * loop is n=2^64-1 base=2^31+1. That gives a | 
 | 112 | 			 * maximum for q=(2^64-1)/2^31 = 0x1ffffffff. Since | 
 | 113 | 			 * base >= 2^31 the loop is finished after a maximum | 
 | 114 | 			 * of three iterations. | 
 | 115 | 			 */ | 
 | 116 | 			while (rx < 0) { | 
 | 117 | 				rx += base; | 
 | 118 | 				q--; | 
 | 119 | 			} | 
 | 120 | 			r = rx; | 
 | 121 | 		} | 
 | 122 | 		*n = q; | 
 | 123 | 	} | 
 | 124 | 	return r; | 
 | 125 | } | 
 | 126 |  | 
 | 127 | #else /* MARCH_G5 */ | 
 | 128 |  | 
 | 129 | uint32_t __div64_32(uint64_t *n, uint32_t base) | 
 | 130 | { | 
 | 131 | 	register uint32_t reg2 asm("2"); | 
 | 132 | 	register uint32_t reg3 asm("3"); | 
 | 133 | 	uint32_t *words = (uint32_t *) n; | 
 | 134 |  | 
 | 135 | 	reg2 = 0UL; | 
 | 136 | 	reg3 = words[0]; | 
 | 137 | 	asm volatile( | 
 | 138 | 		"	dlr	%0,%2\n" | 
 | 139 | 		: "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" ); | 
 | 140 | 	words[0] = reg3; | 
 | 141 | 	reg3 = words[1]; | 
 | 142 | 	asm volatile( | 
 | 143 | 		"	dlr	%0,%2\n" | 
 | 144 | 		: "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" ); | 
 | 145 | 	words[1] = reg3; | 
 | 146 | 	return reg2; | 
 | 147 | } | 
 | 148 |  | 
 | 149 | #endif /* MARCH_G5 */ | 
 | 150 |  | 
 | 151 | EXPORT_SYMBOL(__div64_32); |