例如:
void fun() { } void hun(std::string) { } int main() { function<void(int)> g = &fun; //This fails as it should in my understanding. function<void(int)> f = std::bind(fun); //This works for reasons unkNown to me function<void(int,std::string)> h = std::bind(hun); //this doesn't work return 0; }
如何绑定函数< void(int)>到一个void()函数.
我可以调用f(1)并获得乐趣().
我想了解如何做到这一点.
进入Microsoft Visual Studio 2012的这个实现让我失去了一大堆不可读的宏.所以我在这里问这个问题.
解决方法
std::function<void(int)> f = std::bind(fun,std::placeholders::_1);
对于对Standardese感兴趣的人:
§20.8.9.1.2[func.bind.bind]
template<class F,class... BoundArgs> *unspecified* bind(F&& f,BoundArgs&&... bound_args);
p3 Returns: A forwarding call wrapper
g
with a weak result type (20.8.2). The effect ofg(u1,u2,...,uM)
shall beINVOKE(fd,v1,v2,vN,result_of<FD cv (V1,V2,VN)>::type)
,where cv represents the cv-qualifiers ofg
and the values and types of the bound argumentsv1,vN
are determined as specified below.p10 The values of the bound arguments
v1,vN
and their corresponding typesV1,VN
depend on the typesTiD
derived from the call tobind
and the cv-qualifiers cv of the call wrapperg
as follows:
- if
TiD
isreference_wrapper<T>
,the argument istid.get()
and its typeVi
isT&
;- if the value of
is_bind_expression<TiD>::value
istrue
,the argument istid(std::forward<Uj>(uj)...)
and its typeVi
isresult_of<TiD cv (Uj...)>::type
;- if the value
j
ofis_placeholder<TiD>::value
is not zero,the argument isstd::forward<Uj>(uj)
and its typeVi
isUj&&
;- otherwise,the value is
tid
and its typeVi
isTiD cv &
.
原文地址:https://www.jb51.cc/c/112024.html
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