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Make destructuring trait reference work
Closes rust-lang#15031.
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// Copyright 2012-2014 The Rust Project Developers. See the COPYRIGHT | ||
// file at the top-level directory of this distribution and at | ||
// http://rust-lang.org/COPYRIGHT. | ||
// | ||
// Licensed under the Apache License, Version 2.0 <LICENSE-APACHE or | ||
// http://www.apache.org/licenses/LICENSE-2.0> or the MIT license | ||
// <LICENSE-MIT or http://opensource.org/licenses/MIT>, at your | ||
// option. This file may not be copied, modified, or distributed | ||
// except according to those terms. | ||
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// The regression test for #15031 to make sure destructuring trait | ||
// reference work properly. | ||
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trait T {} | ||
impl T for int {} | ||
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fn main() { | ||
// For an expression of the form: | ||
// | ||
// let &...&x = &..&SomeTrait; | ||
// | ||
// Say we have n `&` at the left hand and m `&` right hand, then: | ||
// if n < m, we are golden; | ||
// if n == m, it's a derefing non-derefable type error; | ||
// if n > m, it's a type mismatch error. | ||
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// n < m | ||
let &x = &(&1 as &T); | ||
let &x = &&(&1 as &T); | ||
let &&x = &&(&1 as &T); | ||
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// n == m | ||
let &x = &1 as &T; //~ ERROR cannot be dereferenced | ||
let &&x = &(&1 as &T); //~ ERROR cannot be dereferenced | ||
let box x = box 1 as Box<T>; //~ ERROR cannot be dereferenced | ||
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// n > m | ||
let &&x = &1 as &T; //~ ERROR found an `&`-pointer pattern | ||
let &&&x = &(&1 as &T); //~ ERROR found an `&`-pointer pattern | ||
let box box x = box 1 as Box<T>; //~ ERROR found a box pattern | ||
} |
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f87bc6a
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r+
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@bors: retry