IEE 754 total order in standard C++11

Question

According to the IEEE floating point wikipage (on IEEE 754), there is a total order on double-precision floating points (i.e. on C++11 implementations having IEEE-754 floats, like gcc 4.8 on Linux / x86-64). Of course, `operator <` on `double` is often providing a total order, but NaN are known to be exceptions (it is well known folklore that `x != x` is a way of testing if `x`, declared as `double x;` is a NaN). The reason I am asking is that I want to have a.g. `std::set<double>` (actually, a set of JSON-like -or Python like- values) and I would like the set to have some canonical representation (my practical concern is to emit portable JSON -same data, ordered in the same order, both on Linux/x86-64 and e.g. on Linux/ARM, even in weird cases like NaN). I cannot find any simple way to get that total order. I coded ``` // a totally ordering function, // return -1 for less-than, 0 for equal, +1 for greater int mydoublecompare(double x, double y) { if (x==y) return 0; else if (x<y) ret…

Accepted Answer

I would use:

```
int totalcompare(double x, double y) {
    int64_t rx, ry;

memcpy(&rx, &x, sizeof rx);
    memcpy(&ry, &y, sizeof ry);

if (rx == ry) return 0;

if (rx < 0) rx ^= INT64_MAX;
    if (ry < 0) ry ^= INT64_MAX;

if (rx < ry) return -1; else return 1;
 }
```

This makes `0.0` and `-0.0` compare unequal, whereas `if (x==y) return 0;` in your version makes them compare equal, meaning that your version is only a preorder. `NaN` values are above the rest and different NaNs compare different. All values comparable for `<=` should be in the same order for the above relation.

Note: the above function is C. I do not know C++.

IEE 754 total order in standard C++11

Solution

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