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digitalmars.D.learn - Append const to array

reply Yuxuan Shui <yshuiv7 gmail.com> writes:
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?
Sep 20 2016
parent reply Jonathan M Davis via Digitalmars-d-learn writes:
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
Sep 20 2016
parent Yuxuan Shui <yshuiv7 gmail.com> writes:
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.
Sep 20 2016