On 12/29/24 1:58 PM, Renato Athaydes wrote:
> I must be misunderstanding how const types work.
>
> I expected this to be accepted by the compiler:
>
> ```
> void foo(const(char[]) c) {
> const(char[])[2] d;
> d[0] = c;
> }
> ```
>
> But it isn't:
>
> ```
> Error: cannot modify `const` expression `d[0]`
> ```
It's the same issue with the following code:
const(char[]) a;
a = c;
The only difference in your example is the fact that my variable 'a'
happens to be array element for you. They are both const(char[]) and
cannot be modified.
> I thought that it would fail only if the `const` applied to the whole
> type,
That would be true but it's "turtles all the way down": everthing that
is accessed through a 'const' expression would be 'const'.
> as in `const(char[][2])` and that the type in the example meant
> only elements are `const` but not the array itself?!
That's true. But the problem you are having is that you are trying to
change an element that is 'const'.
> However, this kind of thing works with slices:
>
> ```
> void foo(const(char[]) c) {
> const(char[])[] d;
> d ~= c;
> }
> ```
>
> Why is this ok, but not the previous one?
Because now you are not changing any element; you are changing the
array, which is not const. The following will still fail with your new
example:
d[0] = c;
> If I use `const(char[][])` instead, then it does fail as I had
> expected... so I can't see where is the hole in my understanding.
Ali
