Why do "const inout" and "const inout shared" exist?

[ag0aep6g via Digitalmars-d]

On 07/02/2017 08:39 AM, H. S. Teoh via Digitalmars-d wrote:
> On Sun, Jul 02, 2017 at 06:49:39AM +0200, Timon Gehr via Digitalmars-d wrote:
>> On 02.07.2017 06:45, Walter Bright wrote:
>>> On 7/1/2017 9:12 PM, Timon Gehr wrote:
[...]
>>>> const(inout(char))[] foo(bool condition, inout(char)[] chars){
>>>>       if(!condition) return "condition failed!";
>>>>       return chars;
>>>> }
[...]
>>
>> I think the example demonstrate the reason. It either returns the
>> argument or an immutable global. If the argument is immutable, so is
>> the return value, otherwise the return value is const.
[...]
> In your example, inout makes no sense at all. It should be written as:
> 
> 	const(char)[] foo(bool condition, const(char)[] chars)
> 
> because if it returns the argument, then it's const(char)[], and if it
> returns an immutable global, then immutable(char)[] implicitly converts
> to const(char)[].

Timon's example makes perfect sense. I don't know if there's an actual 
need for const(inout), but it enables something you can't do with const:

     string s = foo(true, "bar");

> Using inout doesn't make sense here because you
> cannot assume that the return value is mutable if the parameter is
> mutable -- the function may return immutable instead.

You don't get a mutable result for a mutable argument, but you get an 
immutable result for an immutable argument. That's what const(inout) 
enables.