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