为什么取消对迭代器的引用会使输出流中断？

Question

I'm practicing std::deque , specifically iterators

I tested the end iterator. like this:

    std::deque<const char*> wup {"how","are","you","doing","today?"};
    std::deque<const char*>::iterator it = wup.end();

Then, I want to print it , and I know that the end iterator is empty.

Then I try to use 2 cout statements to print the content of it : one of them before --it and another after --it

But if I write a cout before --it , the last cout does not appear in the output.

Something like this:

std::deque<const char*> wup {"how","are","you","doing","today?"};
std::deque<const char*>::iterator it = wup.end();
std::cout<<"Before --it: "<<*it<<std::endl;// okay, it works good | and finishes console
--it;
std::cout<<"After --it: "<<*it<<std::endl;// I do not know why it does not appear.

Output:

  Before --it: Process returned 0 (0x0) execution time : 0.010 s Press ENTER to continue. 

Where did I make a mistake ?
if dereferencing an iterator is wrong ...
why this case, cuts the output-stream without any exception
Thanks a lot. tested on:
OS: Ubuntu 16.01
compiler: g++ version: 5.3.1
IDE: code::blocks 16.01

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Edit: I found this note in /reference/en/cpp/container/map/erase.html :

The iterator pos must be valid and dereferenceable . Thus the end() iterator (which is valid, but is not dereferencable ) cannot be used as a value for pos.

I think this is real reason for, why ... .

my native language is not English, so excuse me, if you see some mistake

Answer 1

Where did i make a mistake?

This line:

std::cout<<"Before --it: "<<*it<<std::endl;

is de-referencing an iterator that points past the end of the container which is undefined behavior.

Why this case, cuts the output-stream without any exception?

This question comes in 2 parts:

1. Why does it cut the output stream? Because undefined behavior is undefined. You don't know what will happen. There is plenty of information around about this: http://en.cppreference.com/w/cpp/language/ub

2. Why no exception? In order to provide an exception to everything that might cause UB, c++ would have to check every de-reference which would be computationally expensive for very little gain. It is up to the user to use the iterator interface correctly. This is the same in the following example:

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 int a[5];
 a[6] = 10; // Why does c++ allow this when it is clearly a mistake? Because the user is 
            // responsible for writing good code, not the compiler, and certainly
            // not the run-time exception handler.