given the next code:
class A {
// ...
};
class B : public A {
// ...
};
And:
A a;
B b;
A& ab = b;
ab = a;
Will be a slicing in the last line of this code? Why?
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1 Answers
1
There will not be slicing. But there will be called the copy assignment operator for the class A due to the static type of the reference ab.
Consider the following program
#include <iostream>
struct A
{
A & operator =( const A & )
{
std::cout << "A::operator = is called" << std::endl;
return *this;
}
};
struct B : A
{
B & operator =( const B & )
{
std::cout << "B::operator = is called" << std::endl;
return *this;
}
};
int main()
{
A a;
B b;
A &ab = b;
ab = a;
return 0;
}
Its output is
A::operator = is called
You may not reassign a reference such a way that it would refer another object.
Vlad from Moscow
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1@DevSolar The question is not about what is slicing. The question is about whether in this example there will be slicing. – Vlad from Moscow Mar 15 '18 at 11:50
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@VladfromMoscow All the members of `b` that declared in `B` will copy to `ab`? – StackUser Mar 15 '18 at 11:53
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1@StackUser There will not be any copy. ab is a reference to b. There is not created any new object. – Vlad from Moscow Mar 15 '18 at 11:54
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@VladfromMoscow, you duplicated an already existed answer. 5 years old answer: Thus, the line a_ref = b1 will call the assignment operator of A, not that of B. And yours: there will be called the copy assignment operator for the class A due to the static type of the reference ab. – Smit Ycyken Mar 15 '18 at 12:02
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@SmitYcyken Five years ago i was not at SO.:) I am a beginner. – Vlad from Moscow Mar 15 '18 at 12:06
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