c++ -

2019-01-07 18:14发布

问题:

In my class called Mat, I want to have a function which takes another function as a parameter. Right now I have the 4 functions below, but I get an error in when calling print(). The second line gives me an error, but I don't understand why, since the first one works. The only difference is function f is not a member of the class Mat, but f2 is. The failure is: 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
}

Why does this happen?

回答1:

The type of pointer-to-member-function is different from pointer-to-function.

The type of a function is different depending on whether it is an ordinary function or a non-static member function of some class:

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

And

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

Note: if it's a static member function of class Fred, its type is the same as if it were an ordinary function: "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.

More on this Here and here.



回答2:

The problem here is that f2 is a method on Mat, while f is just a free function. You can't call f2 by itself, it needs an instance of Mat to call it on. The easiest way around this might be:

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

The = there will capture this, which is what you need to call f2.