问题
Is it possible to take the address of a function that would be found through ADL?
For example:
template<class T>
void (*get_swap())(T &, T &)
{
return & _________; // how do I take the address of T's swap() function?
}
int main()
{
typedef some_type T;
get_swap<T>();
}
回答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 <typename T>
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<a::b>();
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 <a::swap(a::b&, a::b&)>:
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 <void (*get_swap<a::b>())(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.)
来源：https://stackoverflow.com/questions/18086861/is-it-possible-to-take-the-address-of-an-adl-function