[Issue 11308] New: Don't use Voldemort types for std.process output

[d-bugmail at puremagic.com]

http://d.puremagic.com/issues/show_bug.cgi?id=11308

           Summary: Don't use Voldemort types for std.process output
           Product: D
           Version: D2
          Platform: All
        OS/Version: All
            Status: NEW
          Severity: normal
          Priority: P2
         Component: Phobos
        AssignedTo: nobody at puremagic.com
        ReportedBy: andrej.mitrovich at gmail.com
                CC: k.hara.pg at gmail.com


--- Comment #0 from Andrej Mitrovic <andrej.mitrovich at gmail.com> 2013-10-20 13:02:41 PDT ---
-----
import std.process;

void main()
{
    auto res = executeShell("dmd");
    if (res.status) { }
    if (res.output) { }

    // ok, reuse variable
    res = executeShell("rdmd");
    if (res.status) { }
    if (res.output) { }

    // NG. Even though return types are the same
    res = execute(["dmd"]);
}
-----

$ dmd test.d
> Error: cannot implicitly convert expression (execute(["dmd"], null, cast(Config)0, 4294967295u)) of type ProcessOutput to ProcessOutput

The voldemort situation makes reusing variables impossible, even though the
same exact type is returned (the only difference being the template is
instantiated differently).

As a reduced test-case:

-----
auto foo(T)()
{
    struct S { int x; }
    return S(1);
}

void main()
{
    auto res = foo!int();
    res = foo!int();

    // NG: Error: cannot implicitly convert expression (foo()) of type S to S
    res = foo!double();
}
-----

This is a borderline compiler bug. Kenji I've CC'ed you to see if this is
something worth thinking about, could the compiler be smart enough to deduce
that the two Voldemort types are really the same type?

But even so, using `auto` for these process functions makes it very hard to
figure out what the return type of a function is. We might as well use a proper
non-voldemort struct.

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