I'm wondering if a delegate which theoretically doesn't need to make a heap closure, might allocate anyway.
Somehow I don't want to assume the compiler will do the non-allocating way.

This is the code I wrote:
---

    bool feed(float x, Complex!float[] fftData)
    {
        void processSegment(float[] segment)
        {
            fftData.length = _fftSize; // we do need the delegate context here

            /* do stuff with segment and fftData */
        }

        // _analyzer.feed takes a void delegate(float[]) as parameter
        // will this line allocate?
        return _analyzer.feed(x, &processSegment);
    }

---

Does the last line allocate?

On 23/03/2014 23:01, ponce wrote:
> I'm wondering if a delegate which theoretically doesn't need to make a
> heap closure, might allocate anyway.
> Somehow I don't want to assume the compiler will do the non-allocating way.
>
> This is the code I wrote:
> ---
>
>      bool feed(float x, Complex!float[] fftData)
>      {
>          void processSegment(float[] segment)
>          {
>              fftData.length = _fftSize; // we do need the delegate
> context here
>
>              /* do stuff with segment and fftData */
>          }
>
>          // _analyzer.feed takes a void delegate(float[]) as parameter
>          // will this line allocate?
>          return _analyzer.feed(x, &processSegment);
>      }
>
> ---
>
> Does the last line allocate?

In theory, if you use 'scope void delegate(float[])', it won't allocate. But with scope or not, it's probably best to inspect the assembly output if you need to know for sure. Perhaps using http://d.godbolt.org/ can help you do that.