Ways around using a double as a key in a std set/map

Question

The problem of using doubles as keys in maps/sets is floating point precision.

Some people have suggested adding an epsilon in your compare function, but that means you

Answer 1

Convert all the doubles (where we intended as keys) into integers by multiplying them by the precision factor (e.g. 1e8) and rounding to the nearest integer (int)i+0.5(if i>0), then create a set/map that keys off these integers. When extracting the final values of the keys, divide the ints by the precision factor to get the double value back (albeit rounded).

Instead of dividing by the precision factor to get the doubles back, simply store the double together with the associated value in a struct, and put that struct in the dictionary as the "value" for that integer key. That way, the original double value is still around and can be used for calculations. Just not for the key search.

If, however, you can live with slightly rounded values (due to the fact you simply divide an integer by an epsilon), your suggested approach is already good enough.

As the other answer says, it very much depends on the range of the values. If some are extremely huge and others are extremely small, then your approach to get integer keys won't work. If they are only a few digits apart, then it might.

Answer 2

"Convert all the doubles (where we intended as keys) into integers by multiplying them by the precision factor (e.g. 1e8) and rounding to the nearest integer (int)i+0.5(if i>0), then create a set/map that keys off these integers. When extracting the final values of the keys, divide the ints by the precision factor to get the double value back (albeit rounded)."

I would recommend using integer type keys (e.g. long long) for the map in first place, and trim them for double representation using a fixed precision for division.

But that depends, if you are able to apply fix point math for your actual use case. If you need to cover a wide range of value precisions (like e.g. +-1e-7 - +-1e7), such approach won't work.