c++ call to base class method slices object

问题

I have something like this:

#include <iostream>

class X;

class A {
public:
  virtual void bar(X &x);
};

class B : public A {
public:
};

class X {
public:
  void foo(A &a) { std::cout << "foo A" << std::endl; }
  void foo(B &b) { std::cout << "foo B" << std::endl; }
};

void A::bar(X &x) { x.foo(*this); }

int main(int argc, char **argv) {
  X x;
  B b;
  b.bar(x);
  return 0;
}

Compile it and execute it, you'll have:

# ./a.out
foo A
#

I believe this is because object is sliced when cast to A. How can I avoid this so I get

foo B

without implementing the method in B or using some weirdness like the Curiously recurring template pattern ?

回答1:

No slicing is going on here, because you carefully pass objects by reference; slicing requires manipulating the object by value.

The effect is due to overload resolution, which is done statically (i.e. at compile time). When C++ compiles this member function

void A::bar(X &x) {
    x.foo(*this);
}

it needs to decide, at compile time, which of the two overloads to pick. The decision is simple: the compiler knows that *this is of type A, so it calls void foo(A &a) function.

You cannot get it to work without implementing the same method in B^*, using templates, or implementing your own dispatch scheme with function objects or lambdas.

^* in which case you would end up with a nearly classic C++ implementation of the Visitor Pattern, a technique of implementing Double Dispatch.

来源：https://stackoverflow.com/questions/29488392/c-call-to-base-class-method-slices-object

标签

c++

polymorphism

overloading

object-slicing