In the general case you're right, in this case where the optimization is related to the usage of uninitialized memory I don't think that you're correct. Of cour

In the general case you're right, in this case where the optimization is related to the usage of uninitialized memory I don't think that you're correct. Of course this means that the programmer must 'think like an optimizer' and rewrite is-member(i): return sparse[i] < n && dense[sparse[i]] == i as uint sparse = sparse[i] return sparse < n && dense[sparse] == i otherwise the compiler would generate two load instead of 1.