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>
21 lines
352 B
Lua
21 lines
352 B
Lua
f, a, b = Consts("f, a, b")
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nodes = {}
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function mk_big(num)
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local r
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if num == 0 then
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r = f(a, b)
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else
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r = f(mk_big(num-1), mk_big(num-1))
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end
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return r
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end
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function size(e)
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local r = 0
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e:for_each(function(e, o) assert(e:is_app() or e:is_constant()); r = r + 1 end)
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return r
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end
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local F = mk_big(14)
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print(size(F))
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