沒有使用函數指針參數的隱式上傳嗎？

Question

這是我遇到的一種奇怪的情況。我預期的指針是隱含向上轉型：

struct BaseClass 
{}; 

struct DerivedClass : BaseClass 
{}; 

void baseClassArgFunc(BaseClass* arg) {}   // Function taking BaseClass argument 
void derivedClassArgFunc(DerivedClass* arg) {}  // Function taking DerivedClass argument 

int main() 
{ 
    void (*pBaseFuncArg) (BaseClass*);  // Pointer to function taking BaseClass argument 
    void (*pDerivedFuncArg) (DerivedClass*); // Pointer to function taking DerivedClass argument 

    pBaseFuncArg = baseClassArgFunc;   // Assign pointer, works fine 
    pDerivedFuncArg = derivedClassArgFunc;  // Assign pointer, works fine 

    pBaseFuncArg = derivedClassArgFunc;   // A value of type void (*) (DerivedClass *arg) cannot be 
               // assigned to an entity of type void (*) (BaseClass *) 

    pDerivedFuncArg = baseClassArgFunc;   // A value of type void (*) (BaseClass *arg) cannot be 
               // assigned to an entity of type void (*) (DerivedClass *) 

    return 0; 
} 

我預計分配無效（*）（DerivedClass *）爲void baseClassArgFunc（BaseClass的* ARG）是安全的。我很困惑。我想沒有隱式的函數指針參數的上傳嗎？

Answer 1

改變你的類暴露的問題：

struct BaseClass 
{ 
}; 

struct DerivedClass : BaseClass 
{ 
    int a; 
}; 

然後

pBaseFuncArg = derivedClassArgFunc; 

是不安全的：

void derivedClassArgFunc(DerivedClass* arg) { arg->a = 42; } 


int main() 
{ 
    void (*pBaseFuncArg) (BaseClass*) = derivedClassArgFunc; 
    BaseClass base; 

    //derivedClassArgFunc(&base); // Doesn't compile as expected 
    pBaseFuncArg(&base);   // Would be UB, base->i won't exist 
}