rust/src/test/compile-fail/destructure-trait-ref.rs
Niko Matsakis 9e3d0b002a librustc: Remove the fallback to int from typechecking.
This breaks a fair amount of code. The typical patterns are:

* `for _ in range(0, 10)`: change to `for _ in range(0u, 10)`;

* `println!("{}", 3)`: change to `println!("{}", 3i)`;

* `[1, 2, 3].len()`: change to `[1i, 2, 3].len()`.

RFC #30. Closes #6023.

[breaking-change]
2014-06-24 17:18:48 -07:00

42 lines
1.4 KiB
Rust

// 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.
// The regression test for #15031 to make sure destructuring trait
// reference work properly.
trait T {}
impl T for int {}
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.
// n < m
let &x = &(&1i as &T);
let &x = &&(&1i as &T);
let &&x = &&(&1i as &T);
// n == m
let &x = &1i as &T; //~ ERROR cannot be dereferenced
let &&x = &(&1i as &T); //~ ERROR cannot be dereferenced
let box x = box 1i as Box<T>; //~ ERROR cannot be dereferenced
// n > m
let &&x = &1i as &T; //~ ERROR found an `&`-pointer pattern
let &&&x = &(&1i as &T); //~ ERROR found an `&`-pointer pattern
let box box x = box 1i as Box<T>; //~ ERROR found a box pattern
}