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>
25 lines
745 B
C++
25 lines
745 B
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 <unordered_map>
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#include "kernel/expr.h"
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namespace lean {
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template<typename T>
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using expr_map = typename std::unordered_map<expr, T, expr_hash_alloc, expr_eqp>;
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template<typename T>
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using expr_offset_map = typename std::unordered_map<expr_offset, T, expr_offset_hash, expr_offset_eqp>;
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template<typename T>
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using expr_cell_map = typename std::unordered_map<expr_cell *, T, expr_cell_hash, expr_cell_eqp>;
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template<typename T>
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using expr_cell_offset_map = typename std::unordered_map<expr_cell_offset, T, expr_cell_offset_hash, expr_cell_offset_eqp>;
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};
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