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Question

Does the compiler generated assignment operator guard against self assignment?

class T {

   int x;
public:
   T(int X = 0): x(X) {}
};

int main()
{
   T a(1);
   a = a;
}

Do I always need to protect against self-assignment even when the class members aren't of pointer type?

Answer 1

Does the compiler generated assignment operator guard against self assignment?

No, it does not. It merely performs a member-wise copy, where each member is copied by its own assignment operator (which may also be programmer-declared or compiler-generated).

Do I always need to protect against self-assignment even when the class members aren't of pointer type?

No, you do not if all of your class's attributes (and therefore theirs) are POD-types .

When writing your own assignment operators you may wish to check for self-assignment if you want to future-proof your class, even if they don't contain any pointers, et cetera . Also consider the copy-and-swap idiom .

Answer 2

This is an easy one to check empirically:

#include <iostream>
struct A {
  void operator=(const A& rhs) {
    if(this==&rhs) std::cout << "Self-assigned\n";
  }
};

struct B {
  A a;
};

int main()
{
  B b;
  b = b;
}

Answer 3

class T {
    int x;
public:
    T(int X = 0): x(X) {}
// prevent copying
private:
    T& operator=(const T&);
};