691893258d
The new hash code has the property that given expr_cell * c1 and expr_cell * c2, if c1 != c2 then there is a high propbability that c1->hash_alloc() != c2->hash_alloc(). The structural hash code hash() does not have this property because we may have c1 != c2, but c1 and c2 are structurally equal. The new hash code is only compatible with pointer equality. By compatible we mean, if c1 == c2, then c1->hash_alloc() == c2->hash_alloc(). This property is obvious because hash_alloc() does not have side-effects. The test tests/lua/big.lua exposes the problem fixed by this commit. Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
30 lines
1.2 KiB
C++
30 lines
1.2 KiB
C++
/*
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Copyright (c) 2013 Microsoft Corporation. All rights reserved.
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Released under Apache 2.0 license as described in the file LICENSE.
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Author: Leonardo de Moura
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*/
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#pragma once
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#include <utility>
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#include "util/hash.h"
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#include "kernel/expr.h"
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namespace lean {
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typedef std::pair<expr, expr> expr_pair;
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/** \brief Functional object for hashing expression pairs. */
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struct expr_pair_hash {
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unsigned operator()(expr_pair const & p) const { return hash(p.first.hash(), p.second.hash()); }
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};
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/** \brief Functional object for hashing expression pairs (based on their allocation time). */
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struct expr_pair_hash_alloc {
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unsigned operator()(expr_pair const & p) const { return hash(p.first.hash_alloc(), p.second.hash_alloc()); }
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};
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/** \brief Functional object for comparing expression pairs. */
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struct expr_pair_eq {
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unsigned operator()(expr_pair const & p1, expr_pair const & p2) const { return p1.first == p2.first && p1.second == p2.second; }
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};
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/** \brief Functional object for comparing expression pairs using pointer equality. */
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struct expr_pair_eqp {
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unsigned operator()(expr_pair const & p1, expr_pair const & p2) const { return is_eqp(p1.first, p2.first) && is_eqp(p1.second, p2.second); }
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};
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}
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