const types can't be specialized non-const, if arrays?

[Steven Schveighoffer via Digitalmars-d-learn]

On 6/16/16 3:43 PM, cy wrote:
> I don't get it. Do I have to write a separate template for arrays
> specifically or something?
>
> NonConst foo(Constant: const NonConst, NonConst)(Constant bar) {
>     pragma(msg,"NonConst is ",NonConst);
>     pragma(msg,"Constant is ",Constant);
>     NonConst foo = bar;
>     return foo;
> }
>
> void main() {
>     const int bar = 42;
>     auto baz = foo(bar);
>     pragma(msg,typeof(baz));
>     baz = 23;
>     const(int[]) barr = [1,2,3];
>     auto bazz = foo(barr);
>     pragma(msg,typeof(bazz));
>     bazz[0] = 4;
>
> }
>
> /*
>
> NonConst is int
> Constant is const(int)
> int
> NonConst is const(int)[]
> Constant is const(int)[]
> const(int)[]
> derp.d(16): Error: cannot modify const expression bazz[0]
> Failed: ["dmd", "-v", "-o-", "derp.d", "-I."]
>
> */
>

1. Because const(int)[] is not the same as const(int[]). The former is a 
*mutable* array of *constant* integers.
2. Yes, I see that you passed in const(int[]). IFTI was changed a while 
back to automatically convert to tail-const versions when implying 
parameters (to save on template bloat). So it deduces "Constant" to be 
const(int)[].

Also, note that you cannot assign a const item containing references to 
a non-const version of it. So even if NonConst was properly deduced, the 
function would fail to compile.

-Steve