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/*
* Copyright (c) 2003-2005 The Regents of The University of Michigan
* All rights reserved.
*
* Redistribution and use in source and binary forms, with or without
* modification, are permitted provided that the following conditions are
* met: redistributions of source code must retain the above copyright
* notice, this list of conditions and the following disclaimer;
* redistributions in binary form must reproduce the above copyright
* notice, this list of conditions and the following disclaimer in the
* documentation and/or other materials provided with the distribution;
* neither the name of the copyright holders nor the names of its
* contributors may be used to endorse or promote products derived from
* this software without specific prior written permission.
*
* THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
* "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT
* LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR
* A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT
* OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
* SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT
* LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
* DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
* THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
* (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
* OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
*/
#ifndef __BASE_CONDCODE_HH__
#define __BASE_CONDCODE_HH__
#include "base/bitfield.hh"
/**
* Calculate the carry flag from an addition. This should work even when
* a carry value is also added in.
*
* Parameters:
* dest: The result value of the addition.
* src1: One of the addends that was added.
* src2: The other addend that was added in.
*
* Rationale:
* This code analyzes the most sig. bits of the source addends and result,
* and deduces the carry out flag from them without needing the carry in bit.
*
* Observe that we have four cases after an addition regarding the carry
* in and carry out bits:
*
* If we have no carry in but a carry out:
* src1 and src2 must both be 1, with the result bit being 0. Hence,
* ~0 + 1 + 1 => 11, which has a high second bit. We return true.
*
* If we have a carry in and a carry out:
* src1 and src2 can either be 1 and 0, or vice versa. In this case,
* the addition with the carry in gives a result bit of 0 but a carry out.
* Hence,
* ~0 + 1 + 0 => 10, or ~0 + 0 + 1 => 10. We return true.
*
* Or, src1 and src2 can both be one. Along with the carry, this gives
* a result of 1 and a carry out of 1. Hence,
* ~1 + 1 + 1 => 10. We return true.
*
* If we have no carry in and no carry out:
* src1 and src2 can either be 1 and 0, 0 and 1, or 0 and 0.
* In the first two cases the result bit is 1, which when negated does not
* contribute to the sum algorithm at all. In the last case the result bit
* is zero, but neither src1 nor src2 contribute to the sum either. Hence,
* ~1 + 1 + 0 => 1,
* ~1 + 0 + 1 => 1,
* ~0 + 0 + 0 => 1.
* So we return false for all of these cases.
*
* If we have a carry in, but no carry out:
* src1 and src2 can neither be 1. So the overall result bit is 1. Hence:
* ~1 + 0 + 0 => 0. We return false.
*
* @ingroup api_base_utils
*/
static inline bool
findCarry(int width, uint64_t dest, uint64_t src1, uint64_t src2)
{
int shift = width - 1;
return ((~(dest >> shift) & 1) +
((src1 >> shift) & 1) +
((src2 >> shift) & 1)) & 0x2;
}
/**
* Calculate the overflow flag from an addition.
*
* @ingroup api_base_utils
*/
static inline bool
findOverflow(int width, uint64_t dest, uint64_t src1, uint64_t src2)
{
int shift = width - 1;
return ((src1 ^ ~src2) & (src1 ^ dest)) & (1ULL << shift);
}
/**
* Calculate the parity of a value. 1 is for odd parity and 0 is for even.
*
* Parameters:
* dest: a value to be tested.
*
* Rationale:
* findParity simply performs bitwise XOR operations on each "pair" of bits
* in the dest parameter; the procedure being that a pair of ones will be
* XOR'ed out of the intermediate value.
*
* This process is repeated until one last pair of bits are XOR'ed together.
* If the intermediate is still one, then there is exactly one high bit
* which does not have a corresponding high bit. Therefore, the value must
* have odd parity, and we return 1 accordingly. Otherwise we return 0.
*
* @ingroup api_base_utils
*/
static inline bool
findParity(int width, uint64_t dest)
{
dest &= mask(width);
dest ^= (dest >> 32);
dest ^= (dest >> 16);
dest ^= (dest >> 8);
dest ^= (dest >> 4);
dest ^= (dest >> 2);
dest ^= (dest >> 1);
return dest & 1;
}
/**
* Calculate the negative flag.
*
* @ingroup api_base_utils
*/
static inline bool
findNegative(int width, uint64_t dest)
{
return bits(dest, width - 1);
}
/**
* Calculate the zero flag.
*
* @ingroup api_base_utils
*/
static inline bool
findZero(int width, uint64_t dest)
{
return !(dest & mask(width));
}
#endif // __BASE_CONDCODE_HH__