My method, first builds an approximation of the solution of a/b as...

s1 = ((a / 2) / b) * 2;
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a / 2 is done using the bug-free shift-right operator (>>) and results always in a number with bit 64 unset, so we can divide it by b without incurring in the convert-to-NV bug. Again multiplying by 2 is done with the shift-left operator (<<).

Then, we use the approximation to build the exact solution as:

s = s1 + (a - s1 * b) / b
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Well, actually, in order to do everything using bug-free operations, the code uses the following equivalence:

s2 = (a / 2) / b
s1 = s2 * 2
s = s1 + (a - s1 * b) / b
s = s2 * 2 + (a - (s2 * b) * 2) / b
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where $d is s2, and $e is (a - (s2 * b) * 2).