声明一个具有类构造函数作为函数参数的函数

Question

struct A{ 
    A(){} 
}; 

struct B{ 
    B(const A& a){} 
}; 

int main() 
{ 
//Originally I would like to write down below code 

A a; 
B b(a); 

//Unfortunately I end up with below code by accident and there is no compile error 
//I have figured out the below does not create temporary A and call B constructor to 
//create B as the above codes, 
//it declares a function with return value B, name b, 
//and some input parameter, my question is 1) what the input parameter is ? 
//2) How to implement such a function. 

B b(A()); // There is no global function A() in my test case. 

} 

问题是在评论中，我希望有人能帮助我理解它。非常感谢你。

Answer 1

它声明了一个名为b的函数，它返回B，它具有A (*)()类型的单个参数，即指向不带参数的函数的指针并返回A。声明者A()的意思是“不带参数的函数并返回A”，但是无论何时声明具有函数类型的参数，它都会被重写为函数指针。此声明中的参数未命名（如果不想为参数指定名称，则不需要）。

要实施这样的功能，你需要一个定义，例如，，

B b(A a()) { 
    // do something with "a" 
    // note: the type of "a" is still pointer to function 
} 

见，例如，Is there any use for local function declarations?

Answer 2

B b(A())声明了一个名为b的函数，它返回B并将函数指针作为参数。函数指针指向一个返回A并且不带参数的函数。