Clang，std :: shared_ptr和std :: less / operator< [英] Clang, std::shared_ptr and std::less/operator<

问题描述

拥有以下代码

  #include< memory> 
 
 int main（）{
 std :: shared_ptr< int> ptr0（new int）; 
 std :: shared_ptr< int> ptr1（new int）; 
 
 bool result = ptr0< ptr1;使用clang（版本3.1，LLVM 3.1，LLVM 3.1）编译时，会产生以下错误：
} 
  

$ b < Debian GNU / Linux Sid）

  / usr / bin /../ lib / gcc / x86_64-linux-gnu / ../../../../include/c++/4.7/bits/shared_ptr.h:364:14：error：没有匹配的函数调用类型'std :: less< _CT>'$ b的对象$ b return std :: less< _CT>（）（__ a.get（），__b.get（））; 
 ^ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ int>'request here 
 bool result = ptr0< ptr1; 
 ^ 
 /usr/bin/../lib/gcc/x86_64-linux-gnu/4.7/../../../../include/c++/4.7/bits/stl_function .h：236：7：注意：候选函数不可行：对于
，没有从'int *'到'int *&&&'的已知转换。 
 operator（）（const _Tp& __x，const _Tp& __y）const 
 ^ 
  

使用GCC（版本4.7.0）编译相同的代码不会抛出任何错误消息。有没有理由为什么操作符<（）不适用于clang中的共享指针？

解决方案

clang ++和libstdc ++匹配完美。您可以执行以下操作之一：

• 切换到libc ++（使用 clang ++ -stdlib = libc ++ -std = c ++ 11 ... ）



• 将以下补丁应用于 / usr / include / c ++ / 4.7.0 / type_traits （如 http://clang.llvm.org/cxx_status中所述。 html ）：

   索引：include / std / type_traits 
   = ==================================================== ============== 
   --- include / std / type_traits（revision 185724）
   +++ include / std / type_traits（working copy）
   @@ -1746,7 +1746,7 @@ 
   
  模板< typename _Tp，typename _Up> 
   struct common_type< _Tp，_Up> 
    -  {typedef decltype（true？declval< _Tp>（）：declval< _Up>（））type; }; 
   + {typedef typename decay< decltype（true？declat< _Tp>（）：declval< _Up>（））> :: type type; }; 
   
   template< typename _Tp，typename _Up，typename ... _V​​p> 
   struct common_type< _Tp，_Up，_Vp ...> 
    

< bits / shared_ptr.h 你找到了一个 std :: common_type ，clang开发者声称它实际上是libstdc ++的错误，虽然我不相信libstdc ++的错误会导致不存在的类型 int *&&&& 。

Having the following code

#include <memory>

int main() {
    std::shared_ptr<int> ptr0( new int );
    std::shared_ptr<int> ptr1( new int );

    bool result = ptr0 < ptr1;
}

produces the following error when being compiled with clang (version 3.1, LLVM 3.1, Debian GNU/Linux Sid)

/usr/bin/../lib/gcc/x86_64-linux-gnu/4.7/../../../../include/c++/4.7/bits/shared_ptr.h:364:14: error: no matching function for call to object of type 'std::less<_CT>'
      return std::less<_CT>()(__a.get(), __b.get());
             ^~~~~~~~~~~~~~~~
foo.cpp:9:21: note: in instantiation of function template specialization 'std::operator<<int, int>' requested here
        bool result = ptr0 < ptr1;
                           ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/4.7/../../../../include/c++/4.7/bits/stl_function.h:236:7: note: candidate function not viable: no known conversion from 'int *' to 'int *&&&' for
      1st argument;
      operator()(const _Tp& __x, const _Tp& __y) const
      ^

Compiling the same code with GCC (version 4.7.0) doesn't throw any error messages. Is there a reason why operator<() doesn't work for shared pointers in clang?

解决方案

clang++ and libstdc++ doesn't match perfectly yet. You could do one of the followings:

• Switch to libc++ (by using clang++ -stdlib=libc++ -std=c++11 ...)

• Apply the following patch to /usr/include/c++/4.7.0/type_traits (as documented in http://clang.llvm.org/cxx_status.html):

  Index: include/std/type_traits
  ===================================================================
  --- include/std/type_traits (revision 185724)
  +++ include/std/type_traits (working copy)
  @@ -1746,7 +1746,7 @@
  
     template<typename _Tp, typename _Up>
       struct common_type<_Tp, _Up>
  -    { typedef decltype(true ? declval<_Tp>() : declval<_Up>()) type; };
  +    { typedef typename decay<decltype(true ? declval<_Tp>() : declval<_Up>())>::type type; };
  
     template<typename _Tp, typename _Up, typename... _Vp>
       struct common_type<_Tp, _Up, _Vp...>
  

If you check bits/shared_ptr.h you did find a std::common_type, and the clang developers claim that it's actually a bug of libstdc++, although I don't believe a bug of libstdc++ alone would cause the non-existent type int*&&& to appear.

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