2012-12-10 19:32:48 -06:00
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// Copyright 2012 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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(Almost) Always unify a function tail expr with the function result type
typeck::check_fn had an exception for the case where the tail expr
was compatible with type nil -- in that case, it doesn't unify the
tail expr's type with the enclosing function's result type. This
seems wrong to me. There are several test cases in Issue #719
that illustrate why. If the tail expr has type T, for some type
variable T that isn't resolved when this check happens, then T
never gets unified with anything, which is incorrect -- T should
be unified with the result type of the enclosing function. (The
bug was occurring because an unconstrained type variable is
compatible with type nil.)
Instead, I removed the check for type nil and added a check that
the function isn't an iterator -- if it's an iterator, I don't
check the tail expr's type against the function result type,
as that wouldn't make sense.
However, this broke two test cases, and after discussion with
brson, I understood that the purpose of the check was to allow
semicolons to be omitted in some cases. The whole thing seems
rather ad hoc. But I came up with a hacky compromise solution:
instead of checking whether the tailexpr type is *compatible*
with nil, we now just check whether it *is* nil. This also
necessitates calling resolve_type_vars_if_possible before
the check happens, which worries me. But, this fixes the bug
from Issue #719 without requiring changes to any test cases.
Closes #719 but I didn't try every variation -- so reopen the bug
if one of the variations still doesn't work.
2011-08-04 20:34:05 -05:00
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// Tests that the tail expr in null() has its type
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// unified with the type *T, and so the type variable
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// in that type gets resolved.
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2012-09-11 19:46:20 -05:00
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extern mod std;
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(Almost) Always unify a function tail expr with the function result type
typeck::check_fn had an exception for the case where the tail expr
was compatible with type nil -- in that case, it doesn't unify the
tail expr's type with the enclosing function's result type. This
seems wrong to me. There are several test cases in Issue #719
that illustrate why. If the tail expr has type T, for some type
variable T that isn't resolved when this check happens, then T
never gets unified with anything, which is incorrect -- T should
be unified with the result type of the enclosing function. (The
bug was occurring because an unconstrained type variable is
compatible with type nil.)
Instead, I removed the check for type nil and added a check that
the function isn't an iterator -- if it's an iterator, I don't
check the tail expr's type against the function result type,
as that wouldn't make sense.
However, this broke two test cases, and after discussion with
brson, I understood that the purpose of the check was to allow
semicolons to be omitted in some cases. The whole thing seems
rather ad hoc. But I came up with a hacky compromise solution:
instead of checking whether the tailexpr type is *compatible*
with nil, we now just check whether it *is* nil. This also
necessitates calling resolve_type_vars_if_possible before
the check happens, which worries me. But, this fixes the bug
from Issue #719 without requiring changes to any test cases.
Closes #719 but I didn't try every variation -- so reopen the bug
if one of the variations still doesn't work.
2011-08-04 20:34:05 -05:00
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2013-01-23 13:43:58 -06:00
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fn null<T>() -> *T {
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unsafe {
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cast::reinterpret_cast(&0)
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}
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}
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(Almost) Always unify a function tail expr with the function result type
typeck::check_fn had an exception for the case where the tail expr
was compatible with type nil -- in that case, it doesn't unify the
tail expr's type with the enclosing function's result type. This
seems wrong to me. There are several test cases in Issue #719
that illustrate why. If the tail expr has type T, for some type
variable T that isn't resolved when this check happens, then T
never gets unified with anything, which is incorrect -- T should
be unified with the result type of the enclosing function. (The
bug was occurring because an unconstrained type variable is
compatible with type nil.)
Instead, I removed the check for type nil and added a check that
the function isn't an iterator -- if it's an iterator, I don't
check the tail expr's type against the function result type,
as that wouldn't make sense.
However, this broke two test cases, and after discussion with
brson, I understood that the purpose of the check was to allow
semicolons to be omitted in some cases. The whole thing seems
rather ad hoc. But I came up with a hacky compromise solution:
instead of checking whether the tailexpr type is *compatible*
with nil, we now just check whether it *is* nil. This also
necessitates calling resolve_type_vars_if_possible before
the check happens, which worries me. But, this fixes the bug
from Issue #719 without requiring changes to any test cases.
Closes #719 but I didn't try every variation -- so reopen the bug
if one of the variations still doesn't work.
2011-08-04 20:34:05 -05:00
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2011-08-19 17:16:48 -05:00
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fn main() { null::<int>(); }
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