为什么是'X x; X();'允许,当'X'定义一个转换为函数指针,但不是,当它定义一个转换为函子? [英] Why is 'X x; x();' allowed, when 'X' defines a conversion to function pointer, but not, when it defines a conversion to a functor?
问题描述
void f(int){}
typedef void (*f_ptr)(int);
struct Functor{
void operator()(int){}
};
struct X{
operator f_ptr(){ return f; }
};
struct Y{
operator Functor(){ return Functor(); }
};
int main(){
X x; Y y;
x(5); // works ?!
y(5); // doesn't ?!
}
Ideone上的实例。输出:
错误:无法调用'(Y)(int)'
error: no match for call to '(Y) (int)'
Q1:为什么允许调用 x(5)
,即使 X
只定义转换为函数指针,而不是 operator()?
Q1: Why is the call to x(5)
allowed, even though X
only defines a conversion to function pointer, and not operator()
?
Q2:相反,为什么同样的东西不允许,如果我们定义一个转换到另一个函子?
推荐答案
x(5); // works ?!
这会隐式地将 x
转换为 f_ptr
并调用它。 C ++ 11标准:
This implicitly casts x
to an f_ptr
and calls that. C++11 standard:
§13.3.1.1.2调用类类型的对象[over.call.object]
2)此外,对于在T中以
§ 13.3.1.1.2 Call to object of class type [over.call.object]
2) In addition, for each non-explicit conversion function declared in T of the form
operator conversion-type-id ( ) attribute-specifier-seqopt cv-qualifier ;
[...其中 conversion-type-id
表示(P1,...,Pn)的函数的类型指针
返回 R
...] p>
[…where conversion-type-id
denotes the type "pointer to function of (P1,...,Pn)
returning R
"…]
y(5); // doesn't ?!
标准没有提及任何关于重载的类类型的隐式转换,
The standard doesn't mention anything about implicit conversion to class types that overload operator()
(aka functors), which implies that the compiler doesn't allow that.
你必须显式地强制转换:
You must cast it explicitly:
static_cast<Functor>(y)(5);
这篇关于为什么是'X x; X();'允许,当'X'定义一个转换为函数指针,但不是,当它定义一个转换为函子?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!