如何避免显式转换std :: bind()临时对象? [英] How to avoid explicit cast with std::bind() temporary objects?
问题描述
std :: bind的返回类型(有意)未指定。
The return type of std::bind is (intentionally) unspecified. It is storable in a std::function.
下面的示例程序显示了如何显式转换由std返回的临时对象: :bind()到std ::函数,以便调用fn1()。
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().
如果std :: bind的返回类型是可知的,我可以重载回调构造函数&将不再需要显式转换std :: bind临时对象。
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.
有没有办法避免显式转换?
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
}
可能不相关, m在
推荐答案
最好给回调
一个通用构造函数:
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))
{ }
};
(我添加了第二个默认的模板参数,只为类型 T
,语句是有意义的,因此不会创建虚假的可转换属性。)注意这种技术是如何从转换链中移动一个隐式的用户定义的转换:调用显示 con struc tor for fn _
。
(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_
.
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