Use alignof(T) instead of Dune::AlignmentOf<T>::value

The alignof() operator was introduced in C++11.

dune/common/poolallocator.hh

@@ -7,7 +7,6 @@
  * \brief An stl-compliant pool allocator
  */
 
-#include "alignment.hh"
 #include "lcm.hh"
 #include <typeinfo>
 #include <iostream>
@@ -124,7 +123,7 @@ namespace Dune
        * @brief The alignment that suits both the MemberType and
        * the Reference (i.e. their least common multiple).
        */
-      alignment = Lcm<AlignmentOf<MemberType>::value,AlignmentOf<Reference>::value>::value,
+      alignment = Lcm<alignof(MemberType), alignof(Reference)>::value,
 
       /**
        * @brief The aligned size of the type.

dune/common/test/poolallocatortest.cc

@@ -5,7 +5,6 @@
 #endif
 
 #include <dune/common/poolallocator.hh>
-#include <dune/common/alignment.hh>
 #include <dune/common/fmatrix.hh>
 
 using namespace Dune;
@@ -108,7 +107,7 @@ int testPool()
   int ret=0;
 
   std::cout<<"Checking "<<typeid(T).name()<<" sizeof="<<sizeof(T)<<" with size "<< size<<
-  " alignment="<<AlignmentOf<T>::value<<std::endl;
+  " alignment="<< alignof(T) <<std::endl;
 
   ret += testPoolMain<0,T>::test();
   ret += testPoolMain<size,T>::test();
@@ -157,7 +156,7 @@ int main(int, char **)
 
   ret+=testPoolAllocator();
 
-  std::cout<<AlignmentOf<UnAligned>::value<<" "<<sizeof(UnAligned)<<std::endl;
+  std::cout<< alignof(UnAligned) <<" "<<sizeof(UnAligned)<<std::endl;
 
   ret += testPool<UnAligned>();