Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
313 views
in Technique[技术] by (71.8m points)

c++ - const overloaded operator[] function and its invocation

I define two versions of overloaded operator[] function in a class array. ptr is a pointer to first element of the array object.

int& array::operator[] (int sub) {  
   return ptr[sub];
} 

and

int array::operator[] (int sub) const { 
  return ptr[sub];
}

Now, if I define a const object integer1 the second function can only be called..... but if I make a non-const object and then invoke as below:

cout << "3rd value is" << integer1[2];

which function is called here?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Your const version should return const int& not int, so that the semantics are just the same between the two functions.

Once you've done that, it doesn't matter which one is used. If the const version has to be used because your object has a const context, then it will be... and it won't matter as you're not trying to modify anything. Otherwise, it'll use the non-const version... but with just the same effect.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...