Not so perfect forwarding

Consider the following C++ structs:

struct Foo {
  char b;
  int a;
} __attribute__((packed));

struct Bar {
  Bar(int a) {}

Now consider the following code:

#include <memory>

int main() {
  Foo f;
  auto p = std::make_unique<Bar>(f.a);

It should compile right? Wrong:

/tmp/cppsh-NLnGU.cpp: In function ‘int main()’:
/tmp/cppsh-NLnGU.cpp:14:36: error: cannot bind packed field ‘f.Foo::a’ to ‘int&’
   14 |   auto p = std::make_unique<Bar>(f.a);

What gives? I can take it on faith that references to packed fields are banned – in general unaligned memory access is bad. But Bar::Bar takes an int by value, not by reference.

The answer lies in std::make_unique. std::make_unique uses perfect forwarding, a way to forward function arguments “without changing its lvalue or rvalue characteristics” such as const-ness and rvalue-ness. Its function signature looks like this:

template< class T, class... Args >
unique_ptr<T> make_unique( Args&&... args );

In order to forward the perfectly forward the arguments without extra copies, it tries to take a reference, leaving the final (and required) copy to Bar::Bar. Usually this is good, but in this specific case it breaks our code.

To workaround this issue, we could do:

int main() {
  Foo f;
  int a = f.a;
  auto p = std::make_unique<Bar>(a);


Scott Meyers actually lays out a few more perfect forwarding failure cases. Packed members are not covered, but bitfields are probably close enough semantically.