Float[] * Real is a Float[]?

[Jab]

On Tuesday, 10 December 2019 at 20:47:16 UTC, kinke wrote:
> On Tuesday, 10 December 2019 at 20:35:26 UTC, Ali Çehreli wrote:
>> On 12/10/19 11:26 AM, Adam D. Ruppe wrote:
>>> On Tuesday, 10 December 2019 at 19:22:01 UTC, Jonathan Levi 
>>> wrote:
>>>> Why does this not work?
>>> 
>>> The [] operators work in place. It doesn't create a new 
>>> array, just multiplies the existing elements.
>>
>> Then the a[]*r syntax should not be allowed. We should be 
>> forced to write
>>
>>  a[] *= r;
>>
>> No?
>
> No, because the statement is not correct:
>
> void main()
> {
>     float[3] a = [10,2,4];
>     real b = 5.5;
>     float[3] c = a[] * b;
>     assert(a[0] == 10); // fine, `a` is untouched
> }
>
> The actual reason can be seen by inspecting the lowered AST 
> (-vcg-ast):
>
> float[3] c = arrayOp(c[], a[], cast(float)b);
>
> or alternatively, by inspecting the LLVM IR produced by LDC, or 
> the assembly:
>
> pure nothrow @nogc @trusted float[] 
> core.internal.arrayop.arrayOp!(float[], float[], float, "*", 
> "=").arrayOp(float[], float[], float)
>
> I.e., the array-op is in single precision and the scalar rhs b 
> is cast to a float.

Still doesn't really make sense. The compiler is explicitly 
giving those types, it seems to be a bug.

typeof(float * real) == real

That shouldn't be different for arrays, if it should be different 
for an array then you should have to explicitly cast the real to 
float.