Re: Append const to array

[Yuxuan Shui via Digitalmars-d-learn]

On Tuesday, 20 September 2016 at 22:38:33 UTC, Jonathan M Davis 
wrote:
On Tuesday, September 20, 2016 22:23:08 Yuxuan Shui via 
Digitalmars-d-learn wrote:

struct A {
  ulong[] x;
}
struct B {
  ulong x;
}
void main() {
  B[] b;
  const(B) xx = B(1);
  b ~= xx; // Works

  A[] c;
  const(A) yy = A([1]);
  c ~= yy; // Does not
}

What gives?


const(A) means that the ulong[] inside is const(ulong[]). When 
yy is copied
to be appended to c, it goes from const(A) to A, which means 
that
const(ulong[]) would need to be sliced and and set to ulong[], 
which would
violate const, because it would mean that the last element in c 
could mutate
then elements of its x, which would then mutate the elements in 
yy.


- Jonathan M Davis


That makes sense, thanks.

Re: Append const to array

[Jonathan M Davis via Digitalmars-d-learn]

On Tuesday, September 20, 2016 22:23:08 Yuxuan Shui via Digitalmars-d-learn 
wrote:
> struct A {
>   ulong[] x;
> }
> struct B {
>   ulong x;
> }
> void main() {
>   B[] b;
>   const(B) xx = B(1);
>   b ~= xx; // Works
>
>   A[] c;
>   const(A) yy = A([1]);
>   c ~= yy; // Does not
> }
>
> What gives?

const(A) means that the ulong[] inside is const(ulong[]). When yy is copied
to be appended to c, it goes from const(A) to A, which means that
const(ulong[]) would need to be sliced and and set to ulong[], which would
violate const, because it would mean that the last element in c could mutate
then elements of its x, which would then mutate the elements in yy.

- Jonathan M Davis