What's wrong with this Rust macro?

Question

I'm trying to write a macro that generates a From impl for an enum by wrapping the type in a certain variant.

I've come up with this:

macro_rules! variant_derive_from {
    ($enum:ty:$variant:ident($from:ty)) => {
        impl From<$from> for $enum {
            fn from(thing: $from) -> $enum { return $enum::$variant(thing) }
        }
    };
}

However, whenever I try to actually use this macro I get the following error:

error: expected expression, found `B`
|             fn from(thing: $from) -> $enum { $enum::$variant(thing) }
|                                               ^^^^

I couldn't figure out why this might be happening, so I ran a build with macro tracing on, and the macro apparently expands to the following (dummy types, of course):

impl From < A > for B { fn from ( thing : A ) -> B { B :: AVariant ( thing ) } }

When pasting this directly into the code, it compiles successfully. What gives??

Here's a full example of the error.

Stefan · Accepted Answer · 2017-12-02T09:33:45.270

4

A ty placeholder only works as type, not as a namespace. Using ident instead should work just fine in your example.

ident doesn't accept paths ("foo::Bar") though (and using path also doesn't seem to work in this case); you should be able to workaround it by manually constructing paths like matching this: $enum:ident $(:: $enum_path:ident)* (use it like this: $enum $(:: $enum_path)*).

edited Dec 02 '17 at 09:33

answered Dec 02 '17 at 09:23

Stefan

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Won't this cause an issue if the enum is either generic or not directly `use`d by the current file? – AlphaModder Dec 02 '17 at 09:32
You're right; I thought about it too and just edited my answer :) – Stefan Dec 02 '17 at 09:34
Ah, and i suppose it's possible to construct generics support similarly. – AlphaModder Dec 02 '17 at 09:35
Hm... generics can but lifetimes cannot. Whatever, that seems like a really edge case scenario so I'll ignore it. – AlphaModder Dec 02 '17 at 09:46

score 0 · Answer 2 · answered Sep 17 '22 at 04:06

0

You can simply use <_>:: to disambiguate that you want to use $enum as a type:

<$enum>::$variant(thing)

The macro works perfectly after that change (playground).

answered Sep 17 '22 at 04:06

kmdreko

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What's wrong with this Rust macro?

2 Answers2