问题
The return type of std::bind is (intentionally) unspecified. It is storable in a std::function.
The example program below shows how I have to explicitly cast the temporary object returned by std::bind() to a std::function in order to call fn1().
If the return type of std::bind was knowable, I could overload the Callback constructor & would no longer need to explicitly cast std::bind temporary objects.
Is there any way to avoid the explicit cast?
// g++ -std=c++11 test.cxx
#include <functional>
using std::placeholders::_1;
class A
{
public:
void funcA (int x) { }
};
class Callback
{
public:
Callback () = default;
Callback (std::function<void(int)> f) { }
// Wish we knew the return type of std::bind()
// Callback (return_type_of_std_bind f) { }
};
void fn0 (std::function<void(int)> f) { }
void fn1 (Callback cb) { }
int main (void)
{
A a;
fn0(std::bind(&A::funcA, &a, _1)); // ok
fn1(std::function<void(int)>(std::bind(&A::funcA, &a, _1))); // ok, but verbose
fn1(std::bind(&A::funcA, &a, _1)); // concise, but won't compile
}
Probably not relevant, but I'm using gcc 4.7.2 on Linux.
回答1:
Best to give Callback
a universal constructor:
struct Callback
{
typedef std::function<void(int)> ftype;
ftype fn_;
template <typename T,
typename = typename std::enable_if<std::is_constructible<ftype, T>::value>::type>
Callback(T && f) : fn_(std::forward<T>(f))
{ }
};
(I added the second, defaulted template argument to only enable this constructor for types T
for which the statement makes sense, so as not to create false convertibility properties.) Note how this technique removes one implicit user-defined conversion from the conversion chain, by invoking an explicit constructor for fn_
.
来源:https://stackoverflow.com/questions/14348681/how-to-avoid-explicit-cast-with-stdbind-temporary-objects