c++ - <未解析的重载函数类型>

c++ - <unresolved overloaded function type>

提问by Mordrag

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 fis not a member of the class Mat, but f2is. The failure is: error: no matching function for call to Mat::test( < unresolved overloaded function type>, int)'

在我的类中Mat，我想要一个函数，它将另一个函数作为参数。现在我有下面的 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，因为第一行有效。唯一的区别是 functionf不是 class 的成员Mat，而是f2。失败是：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
}

Why does this happen?

为什么会发生这种情况？

回答by

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

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

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

函数的类型取决于它是普通函数还是某个类的非静态成员函数：

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

And

和

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)"

注意：如果是类 Fred 的静态成员函数，则其类型与普通函数相同： "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.

在 C++ 中，成员函数有一个指向对象的隐式参数（成员函数内部的 this 指针）。可以认为普通 C 函数具有与成员函数不同的调用约定，因此它们的指针类型（指向成员函数的指针与指向函数的指针）不同且不兼容。C++ 引入了一种新的指针类型，称为指向成员的指针，它只能通过提供一个对象来调用。

注意：不要试图将指向成员函数的指针“强制转换”为指向函数的指针；结果是不确定的，可能是灾难性的。例如，指向成员函数的指针不需要包含适当函数的机器地址。如上一个示例中所述，如果您有一个指向常规 C 函数的指针，请使用顶级（非成员）函数或静态（类）成员函数。

More on this Hereand here.

更多关于这个here和here。

回答by Barry

The problem here is that f2is a method on Mat, while fis just a free function. You can't call f2by itself, it needs an instance of Matto call it on. The easiest way around this might be:

这里的问题是这f2是一个方法 on Mat，而f只是一个自由函数。你不能自己调用f2，它需要一个实例Mat来调用它。解决此问题的最简单方法可能是：

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

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

在=那里将捕获this，这是你需要叫什么f2。

回答by CourageousPotato

I got this error from not using parentheses for a function call. I had this:

我因为函数调用没有使用括号而得到这个错误。我有这个：

Eigen::Quaterniond value; ... myFunction(value.x);

Eigen::Quaterniond value; ... myFunction(value.x);

And I fixed this by changing the last line to this:

我通过将最后一行更改为以下内容来解决此问题：

myFunction(value.x());

myFunction(value.x());