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?

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

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.