Is this a violation of const?

[H. S. Teoh]

On Fri, Jul 29, 2022 at 09:56:20PM +0000, Andrey Zherikov via Digitalmars-d-learn wrote:
> In the example below `func` changes its `const*` argument. Does this
> violates D's constness?
> 
> ```d
> import std;
> 
> struct S
> {
>     string s;
> 
>     void delegate(string s) update;
> }
> 
> void func(const S* s)
> {
>     writeln(*s);
>     s.update("func");
>     writeln(*s);
> }
> 
> void main()
> {
>     auto s = S("test");
>     s.update = (_) { s.s = _; };
> 
>     writeln(s);
>     func(&s);
>     writeln(s);
> }
> ```
> 
> The output is:
> ```
> S("test", void delegate(string))
> const(S)("test", void delegate(string))
> const(S)("func", void delegate(string))
> S("func", void delegate(string))
> ```

At first I thought this was a bug in the const system, but upon closer
inspection, this is expected behaviour. The reason is, `const`
guarantees no changes *only on the part of the recipient* of the `const`
reference; it does not guarantee that somebody else doesn't have a
mutable reference to the same data.  For the latter, you want immutable
instead of const.

So in this case, func receives a const reference to S, so it cannot
modify S. However, the delegate created by main() *can* modify the data,
because it holds a mutable reference to it. So when func invokes the
delegate, the delegate modifies the data thru its mutable reference.

Had func been declared with an immutable parameter, it would have been a
different story, because you cannot pass a mutable argument to an
immutable parameter, so compilation would fail. Either s was declared
mutable and the delegate can modify it, but you wouldn't be able to pass
it to func(), or s was declared immutable and you can pass it to func(),
but the delegate creation would fail because it cannot modify immutable.

In a nutshell, `const` means "I cannot modify the data", whereas
`immutable` means "nobody can modify the data". Apparently small
difference, but actually very important.


T

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