Is it possible to take the address of an ADL function?

Question

Is it possible to take the address of a function that would be found through ADL?

For example:

template
void (*get_swap())(T &, T &         


        

Answer 1

Honestly, I don't know but I tend towards saying that this is not possible.

Depending on what you want to achieve I can suggest a workaround. More precisely, if you just need the address of a function that has the same semantics as swap called through ADL then you can use this:

template 
void (*get_swap())(T&, T&) {
    return [](T& x, T& y) { return swap(x, y); };
}

For instance, the following code:

namespace a {

  struct b {
      int i;
  };

  void swap(b& x, b& y) {
      std::swap(x.i, y.i);
  }
}

int main() {

    auto f0 = (void (*)(a::b&, a::b&)) a::swap;
    auto f1 = get_swap();

    std::cout << std::hex;
    std::cout << (unsigned long long) f0 << '\n';
    std::cout << (unsigned long long) f1 << '\n';
}

compiled with gcc 4.8.1 (-std=c++11 -O3) on my machine gave:

4008a0
4008b0

The relevant assembly code (objdump -dSC a.out) is

00000000004008a0 :
  4008a0:   8b 07                   mov    (%rdi),%eax
  4008a2:   8b 16                   mov    (%rsi),%edx
  4008a4:   89 17                   mov    %edx,(%rdi)
  4008a6:   89 06                   mov    %eax,(%rsi)
  4008a8:   c3                      retq   
  4008a9:   0f 1f 80 00 00 00 00    nopl   0x0(%rax)

00000000004008b0 ())(a::b&, a::b&)::{lambda(a::b&, a::b&)#1}::_FUN(a::b&, a::b&)>:
  4008b0:   8b 07                   mov    (%rdi),%eax
  4008b2:   8b 16                   mov    (%rsi),%edx
  4008b4:   89 17                   mov    %edx,(%rdi)
  4008b6:   89 06                   mov    %eax,(%rsi)
  4008b8:   c3                      retq   
  4008b9:   0f 1f 80 00 00 00 00    nopl   0x0(%rax)

As one can see the functions pointed by f0 and f1 (located at 0x4008a0 and 0x4008b0, respectively) are binary identical. The same holds when compiled with clang 3.3.

If the linker can do identical COMDAT folding (ICF), I guess, we can even get f0 == f1. (For more on ICF see this post.)