C++ - <无法解析的重载函数类型>

Question

在我的课程Mat，我希望有一个函数，将另一个函数作为参数。现在我有以下4个函数，但是在调用print（）时出现错误。第二行给了我一个错误，但我不明白为什么，因为第一行是有效的。唯一的区别是功能f不是类Mat的成员，但是f2是。 故障是：error: no matching function for call to Mat::test(< unresolved overloaded function type>, int)'

template <typename F> 
int Mat::test(F f, int v){ 
    return f(v); 
} 

int Mat::f2(int x){ 
    return x*x; 
} 

int f(int x){ 
    return x*x; 
} 

void Mat::print(){ 
    printf("%d\n",test(f ,5)); // works 
    printf("%d\n",test(f2 ,5)); // does not work 
} 

为什么会出现这种情况？

Answer 1

这里的问题是f2是Mat上的方法，而f只是一个免费的功能。您不能自己拨打f2，需要拨打Mat的实例才能打电话。解决这个问题的最简单的方法可能是：

printf("%d\n", test([=](int v){return this->f2(v);}, 5)); 

的=会有捕捉this，这是你需要调用f2什么。

Answer 2

pointer-to-member-function的类型不同于pointer-to-function。

取决于它是否是一个普通函数或某些类的非静态成员函数类型的功能是不同的：

int f(int x); 
the type is "int (*)(int)" // since it is an ordinary function 

而且

int Mat::f2(int x); 
the type is "int (Mat::*)(int)" // since it is a non-static member function of class Mat 

注：如果它是Fred类的静态成员函数，其类型与普通函数类型相同："int (*)(char,float)"

In C++, member functions have an implicit parameter which points to the object (the this pointer inside the member function). Normal C functions can be thought of as having a different calling convention from member functions, so the types of their pointers (pointer-to-member-function vs pointer-to-function) are different and incompatible. C++ introduces a new type of pointer, called a pointer-to-member, which can be invoked only by providing an object.

NOTE: do not attempt to "cast" a pointer-to-member-function into a pointer-to-function; the result is undefined and probably disastrous. E.g., a pointer-to-member-function is not required to contain the machine address of the appropriate function. As was said in the last example, if you have a pointer to a regular C function, use either a top-level (non-member) function, or a static (class) member function.

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