On Wednesday, 13 September 2017 at 17:39:29 UTC, Steven
Schveighoffer wrote:
Correct. So given a function:
inout(int*) foo(inout(int*)p1, inout(int*)p2)
The table shows what inout is resolved as when calling the
function.
If you consider the column the mutability of p1, and the row
the mutability of p2, then the value in the table represents
the mutability of the return value.
So for instance:
int *m;
const int *c;
immutable int *i;
inout int *w;
auto v1 = foo(m, m); // typeof(v1) is int*
auto v2 = foo(m, c); // typeof(v2) is const(int*)
auto v3 = foo(i, m); // typeof(v3) is const(int*)
auto v4 = foo(w, w); // typeof(v4) is inout(int*)
auto v5 = foo(w, i); // typeof(v5) is inout(const(int *))
etc.
Thank you. Now it is clear to me. The source of my confusion was
that, say, given a function:
inout(int*) foo(inout(int*) p)
as per table, combining (mutable) argument int* m with parameter
inout(int*) p would produce parameter type const(int*). But now I
see that the table deduces the common type of all parameters, not
the common type of a parameter and its argument (the
documentation is kind of hard to parse: "If such a match occurs,
the inout is considered the common qualifier of the matched
qualifiers...").
