[Issue 16390] __traits not accepted where a type is expected

[via Digitalmars-d-bugs]

https://issues.dlang.org/show_bug.cgi?id=16390

Jonathan M Davis <issues.dlang at jmdavisProg.com> changed:

           What    |Removed                     |Added
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                 CC|                            |issues.dlang at jmdavisProg.co
                   |                            |m

--- Comment #4 from Jonathan M Davis <issues.dlang at jmdavisProg.com> ---
Similarly,

template members(T)
{
    alias members = __traits(allMembers, T);
}

doesn't work, whereas

template members(T)
{
    alias members = AliasSeq!(__traits(allMembers, T));
}

does. And that's even more bizarre when you consider not only that AliasSeq is
supposed to be equivalent to the "tuple" that the compiler uses for stuff like
the result of __traits(allMembers, T), but this code

import std.meta;

template members(T)
{
    alias members = AliasSeq!(__traits(allMembers, T));
}

struct S
{
    int i;
    string s;
}

void main()
{
    pragma(msg, __traits(allMembers, S));
    pragma(msg, AliasSeq!(__traits(allMembers, S)));
    pragma(msg, members!S);
}

prints

tuple("i", "s")
tuple("i", "s")
tuple("i", "s")

meaning that all 3 are equivalent as far as pragma(msg, ...) is concerned. And
changing main to

    pragma(msg, typeof(__traits(allMembers, S)));
    pragma(msg, typeof(AliasSeq!(__traits(allMembers, S))));
    pragma(msg, typeof(members!S));

results in

(string, string)
(string, string)
(string, string)

So, it really seems like the fact that the result of __traits can't be aliased
is a bug. And if it's not a bug, then it's an unnecessary (and confusing)
inconsistency, and IMHO, it really should be fixed.

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