270f0eef73
This is the kind of change that one is expected to need to make to accommodate overloaded-`box`. ---- Note that this is not *all* of the changes necessary to accommodate Issue 22181. It is merely the subset of those cases where there was already a let-binding in place that made it easy to add the necesasry type ascription. (For unnamed intermediate `Box` values, one must go down a different route; `Box::new` is the option that maximizes portability, but has potential inefficiency depending on whether the call is inlined.) ---- There is one place worth note, `run-pass/coerce-match.rs`, where I used an ugly form of `Box<_>` type ascription where I would have preferred to use `Box::new` to accommodate overloaded-`box`. I deliberately did not use `Box::new` here, because that is already done in coerce-match-calls.rs. ---- Precursor for overloaded-`box` and placement-`in`; see Issue 22181.
44 lines
1.4 KiB
Rust
44 lines
1.4 KiB
Rust
// Copyright 2014 The Rust Project Developers. See the COPYRIGHT
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// file at the top-level directory of this distribution and at
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// http://rust-lang.org/COPYRIGHT.
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//
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// Licensed under the Apache License, Version 2.0 <LICENSE-APACHE or
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// http://www.apache.org/licenses/LICENSE-2.0> or the MIT license
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// <LICENSE-MIT or http://opensource.org/licenses/MIT>, at your
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// option. This file may not be copied, modified, or distributed
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// except according to those terms.
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#![allow(unknown_features)]
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#![feature(box_syntax)]
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struct Foo(Box<isize>, isize);
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struct Bar(isize, isize);
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fn main() {
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let x: (Box<_>, _) = (box 1, 2);
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let r = &x.0;
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let y = x; //~ ERROR cannot move out of `x` because it is borrowed
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let mut x = (1, 2);
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let a = &x.0;
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let b = &mut x.0; //~ ERROR cannot borrow `x.0` as mutable because it is also borrowed as
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let mut x = (1, 2);
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let a = &mut x.0;
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let b = &mut x.0; //~ ERROR cannot borrow `x.0` as mutable more than once at a time
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let x = Foo(box 1, 2);
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let r = &x.0;
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let y = x; //~ ERROR cannot move out of `x` because it is borrowed
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let mut x = Bar(1, 2);
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let a = &x.0;
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let b = &mut x.0; //~ ERROR cannot borrow `x.0` as mutable because it is also borrowed as
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let mut x = Bar(1, 2);
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let a = &mut x.0;
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let b = &mut x.0; //~ ERROR cannot borrow `x.0` as mutable more than once at a time
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}
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