Is floating point arithmetic pure?

Calculations are performed in an extended register of 80 bits
The extended representation is used for as long as possible (until you ship
the value to memory)
So the following is possible:

```cpp
double func(); // nothing strange done in func, like modifying global state

assert(func() != func()); // this can pass because one func() 80-bit value
gets stored on stack so truncated, and it will be compared against
nontruncated value
```

On Tue, Mar 4, 2025, 11:59 Ell via Std-Proposals <
std-proposals_at_[hidden]> wrote:

>
>
> On 3/4/25 12:20, Tymi via Std-Proposals wrote:
> > Yes, the compiler might assume that the function produces the same
> > effect for the same arguments.
> int sum (const std::vector<int>& v) [[pure]];
>
> void f () {
> std::vector<int> v = {1, 2, 3};
>
> int s1 = sum (v);
>
> v.push_back (4);
>
> int s2 = sum (v); // is this the same argument?
> }
>
> void g () {
> std::vector<int> v = {1, 2, 3};
> std::vector<int> w = {1, 2, 3};
>
> int s1 = sum (v);
> int s2 = sum (w); // is this the same argument?
> }
> --
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>