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<base href="http://llvm.org/bugs/" />
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<th>Bug ID</th>
<td><a class="bz_bug_link
bz_status_NEW "
title="NEW --- - generic lambdas, decltype(auto), and rvalue references, oh my!"
href="http://llvm.org/bugs/show_bug.cgi?id=22426">22426</a>
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<th>Summary</th>
<td>generic lambdas, decltype(auto), and rvalue references, oh my!
</td>
</tr>
<tr>
<th>Product</th>
<td>clang
</td>
</tr>
<tr>
<th>Version</th>
<td>trunk
</td>
</tr>
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<th>Hardware</th>
<td>PC
</td>
</tr>
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<th>OS</th>
<td>Windows NT
</td>
</tr>
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<th>Status</th>
<td>NEW
</td>
</tr>
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<th>Severity</th>
<td>normal
</td>
</tr>
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<th>Priority</th>
<td>P
</td>
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<th>Component</th>
<td>C++14
</td>
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<th>Assignee</th>
<td>unassignedclangbugs@nondot.org
</td>
</tr>
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<th>Reporter</th>
<td>eniebler@boost.org
</td>
</tr>
<tr>
<th>CC</th>
<td>llvmbugs@cs.uiuc.edu
</td>
</tr>
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<th>Classification</th>
<td>Unclassified
</td>
</tr></table>
<p>
<div>
<pre>Clang (trunk) rejects the following code (with -std=c++1y):
#include <utility>
int main()
{
using std::pair;
using std::declval;
using X = decltype(declval<pair<int&&,int&&>>().first);
auto f = [](auto && p) -> decltype(auto) //((decltype(p)&&)p).first)
{
return ((decltype(p)&&)p).first;
};
using Y = decltype(f(declval<pair<int&&,int&&>>()));
}
The error I get is:
main.cpp:13:16: error: rvalue reference to type 'int' cannot
bind to lvalue of type 'int'
return ((decltype(p)&&)p).first;
^~~~~~~~~~~~~~~~~~~~~~~~
There should *never* be an error when using decltype(auto). The compiler should
use the actual type of the return expression as the return type. It seems clang
is getting confused about the value category here.
I observe that X is an alias for int&& in the above code, and I feel that the
generic lambda is doing the same thing. Both the type of the expression in the
lambda and the return type of the lambda should also be int&&.
I also observe that if I replace decltype(auto) with
decltype(((decltype(p)&&)p).first), then the code compiles and Y is an alias
for int&&, which seems right to me.
For the record, gcc also gets this code wrong, but in a different way.</pre>
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