about the new const regime

[Daniel919]

> c is not a copy of s or a reference to s.  c references the same data that s 
> references.  When you change s to reference different data, it does not 
> change c to reference this data.  i.e. after the assignment, s.ptr != c.ptr. 
> Maybe this example will help:
> 
> int x = 5;
> int *p = &x;
> int *p2 = p;
> writefln(*p, *p2); // outputs 55
> 
> int y = 6;
> p = &y;
> 
> writefln(*p, *p2); // outputs 65
> 
> Assigning p to &y is similar to your code assigning s to "new".dup;
> 
> Now, if you modified a piece of the data s is pointing to, then c will 
> change.  Try this:
> 
> s[0..3] = "new";
> This copies the data from the array on the right into the array on the left 
> (I didn't yet compile this, so I'm not sure if it compiles as coded).
writefln( c ); //test
s[0..3] = "new";
writefln( c ); //newt
This made it clear, shows that c is not a copy of s.

And as expected:
c[0..3] = "new";//Error: slice c[cast(uint)0..cast(uint)3] is not mutable


const(char)[] c = s;
elements c refers to can't be changed via c

invariant(char)[] i = s;
elements i refers to will never change


Thanks for your explanation.