How does `as` break inference?

help
1

We believe constructs of the form:

fn bar<T>(x: *const T, y: something) -> T { ...}

let x = foo as *const _;
bar(x, something)

can break inference, while constructs of the form:

fn bar<T>(x: *const T, y: something) -> T { ...}

let x: *const _ = foo;
bar(x, something)

would have proper inference, but we're struggling to come up with an example where the latter compiles but the former fails to compile. Anyone got any ideas?

2

found it Rust Playground

3

Did you mean to post a different link? The code in the link you posted doesn't use any as expressions, except in the offset calculation in the cell_project!() macro.

4

(pronouns fixed)

note the commented-out let field = field as *mut _;. can't exactly have the 2 examples in the 1 playground...

5

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