問題描述

在我的名為 Mat 的類中，我想要一個將另一個函數作為參數的函數.現在我有下面的 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，但我不明白為什么，因為第一行有效.唯一的區別是函數 f 不是 Mat 類的成員，但 f2 是.失敗是:error: no matching function for call to Mat::test( < unresolved overloaded function type>, int)'

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
",test(f ,5));    // works
    printf("%d
",test(f2 ,5));    // does not work
}

為什么會發生這種情況?

Why does this happen?

推薦答案

pointer-to-member-function 的類型與 pointer-to-function 不同.

函數的類型根據是普通函數還是某個類的非靜態成員函數而不同:

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

還有

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

注意:如果是Fred類的靜態成員函數，其類型與普通函數相同:"int (*)(char,float)"

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)"

在 C++ 中，成員函數有一個隱式參數指向對象(成員函數內的 this 指針).正常 C函數可以被認為具有不同的調用約定來自成員函數，所以它們的指針類型(指向成員函數的指針與指向函數的指針)不同并且不兼容. C++ 引入了一種新的指針類型，稱為成員指針，只能通過提供對象調用.

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.

注意:不要試圖將指向成員函數的指針強制轉換"為函數指針；結果是不確定的，可能是災難性的.例如，指向成員函數的指針不需要包含相應函數的機器地址. 正如上次所說例如，如果您有一個指向常規 C 函數的指針，請使用頂級(非成員)函數，或靜態(類)成員函數.

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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