Confusingly about pop_back (), C ++

Question

I am trying to understand the behavior of vector::pop_back() . Therefore, I have the following code snippet:

 vector<int> test; test.push_back(1); test.pop_back(); cout << test.front() << endl; 

This may be right, but it surprises me that it prints 1. So I'm confused. Does pop_back() only have a way to remove an index > 0 element?

Thanks in advance!

Answer 1

You invoke undefined behavior by calling front on an empty vector. This is similar to indexing the boundaries of an array. Anything can happen, including refund 1 .

Answer 2

pop_back here leaves the vector empty. Access to empty vectors front() is undefined behavior. This means that something can happen - there is no prediction. It may fall, it may return 1, it may return 42, it may print “Hello world!” It may erase your hard drive, or it may cause demons through your nasal passages . Any of these are acceptable actions as far as C ++ is concerned - return 1 just ended here. Do not rely on this.