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>
