C++ - <無法解析的重載函數類型>

Question

在我的課程Mat，我希望有一個函數，將另一個函數作爲參數。現在我有以下4個函數，但是在調用print（）時出現錯誤。第二行給了我一個錯誤，但我不明白爲什麼，因爲第一行是有效的。唯一的區別是功能f不是類Mat的成員，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 
} 

爲什麼會出現這種情況？

Answer 1

這裏的問題是f2是Mat上的方法，而f只是一個免費的功能。您不能自己撥打f2，需要撥打Mat的實例才能打電話。解決這個問題的最簡單的方法可能是：

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

的=會有捕捉this，這是你需要調用f2什麼。

Answer 2

pointer-to-member-function的類型不同於pointer-to-function。

取決於它是否是一個普通函數或某些類的非靜態成員函數類型的功能是不同的：

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類的靜態成員函數，其類型與普通函數類型相同："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.

更多關於Here和here。