6220: implement binary operator overloading type inference r=flodiebold a=ruabmbua
Extend type inference of *binary operator expression*, by adding support for operator overloads.
Before this merge request, the type inference of binary expressions could only resolve operations done on built-in primitive types. This merge requests adds a code path, which is executed in case the built-in inference could not get any results. It resolves the proper operator overload trait in *core::ops* via lang items, and then resolves the associated *Output* type.
```rust
struct V2([f32; 2]);
#[lang = "add"]
pub trait Add<Rhs = Self> {
/// The resulting type after applying the `+` operator.
type Output;
/// Performs the `+` operation.
#[must_use]
fn add(self, rhs: Rhs) -> Self::Output;
}
impl Add<V2> for V2 {
type Output = V2;
fn add(self, rhs: V2) -> V2 {
let x = self.0[0] + rhs.0[0];
let y = self.0[1] + rhs.0[1];
V2([x, y])
}
}
fn test() {
let va = V2([0.0, 1.0]);
let vb = V2([0.0, 1.0]);
let r = va + vb; // This infers to V2 now
}
```
There is a problem with operator overloads, which do not explicitly set the *Rhs* type parameter in the respective impl block.
**Example:**
```rust
impl Add for V2 {
type Output = V2;
fn add(self, rhs: V2) -> V2 {
let x = self.0[0] + rhs.0[0];
let y = self.0[1] + rhs.0[1];
V2([x, y])
}
}
```
In this case, the trait solver does not realize, that the *Rhs* type parameter is actually self in the context of the impl block. This stops type inference in its tracks, and it can not resolve the associated *Output* type.
I guess we can still merge this back, because it increases the amount of resolved types, and does not regress anything (in the tests).
Somewhat blocked by https://github.com/rust-analyzer/rust-analyzer/issues/5685
Resolves https://github.com/rust-analyzer/rust-analyzer/issues/5544
Co-authored-by: Roland Ruckerbauer <roland.rucky@gmail.com>
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