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

[Walter Bright via Digitalmars-d]

On 7/2/2017 6:33 AM, Timon Gehr wrote:
The best way to think about inout is that it enables a function to have three 
distinct signatures:

inout(int)[] foo(inout(int)[] arg);

"expands" to:

int[] foo(int[] arg);
immutable(int)[] foo(immutable(int)[] arg);
const(int)[] foo(const(int)[] arg);


const inout /does not change this in any way/:


const(inout(int))[] foo(inout(int)[] arg);

expands to:

const(int)[] foo(int[] arg);
const(immutable(int))[] foo(immutable(int)[] arg);
const(const(int))[] foo(const(int)[] arg);

Thank you. This explanation makes sense (given that applying const to immutable 
=> immutable).