/* * 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. * * Authors: Gabe Black */ #ifndef __BASE_CONDCODE_HH__ #define __BASE_CONDCODE_HH__ #include "base/bitfield.hh" #include "base/trace.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. */ 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. */ 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. */ 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. */ inline bool findNegative(int width, uint64_t dest) { return bits(dest, width - 1, width - 1); } /** * Calculate the zero flag. */ inline bool findZero(int width, uint64_t dest) { return !(dest & mask(width)); } #endif // __BASE_CONDCODE_HH__