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#include <vector> 
struct Foo; 
 std::vector<Foo> getFilteredFoos(bool (*filter)(const Foo &) = [](const Foo &){return true;});

prog.cpp:3:79: error: type ‘__lambda0’ with no linkage used to declare function ‘auto __lambda0::operator()(const Foo&) const’ with linkage [-fpermissive]
  std::vector<Foo> getFilteredFoos(bool (*filter)(const Foo &) = [](const Foo &){return true;});
                                                                               ^
prog.cpp:3:93: warning: lambda expressions only available with -std=c++11 or -std=gnu++11 [enabled by default]
  std::vector<Foo> getFilteredFoos(bool (*filter)(const Foo &) = [](const Foo &){return true;});
                                                                                             ^

Standard output is empty