解引用C ++ STL列表迭代器

Question

I'm looping through two STL lists (L1, L2) like so:

list<int>::const_iterator itr1 = L1.begin();
list<int>::const_iterator itr2 = L2.begin();

for (itr1; itr1 != L1.end(); itr1++) {
   if (*itr1 < *itr2) {
     //some code
   }

}

It compiles fine but when I run it, it says "Expression: list iterator not dereferencable"

Now in class we made a mock version of the STL list where we wrote our own STL list and we had overloaded the *operator to dereference an iterator. However, obviously it's not working here.

How can I dereference an iterator, or if STL list does it differently, how does it do it. I looked through this:

http://www.sgi.com/tech/stl/List.html

documentation and didn't seem to find anything accept the member "reference" but still did not see how to reference what an iterator is pointing to, unless it's the first or last part of the list.

Anyone know what's going on here? Thank you

here is a pastebin:

http://pastebin.com/YRddqjmN

Answer 1

My guess:

L2 is empty, so L2.begin() is the same as L2.end() .

Which means L2.begin() is returning a non-referencable iterator and you are thus invoking undefined behavior.

Answer 2

This (implementation specific) message suggests to me that you dereferenced an invalid iterator. This has nothing to do with syntax/compile-time semantics so no surprise that your compiler didn't complain. However note that iterators do have run-time semantics: in this case I'd wager that the code is called with an empty L2 list, so that itr2 == L2.end() . That means that *itr2 results in undefined behaviour. Luckily this seems to trigger an error-message rather than blowup in your face.

Answer 3

while ( *itr2 < *itr1 ) {
    itr2++;
}

That code has no check for running off the end of L2. Maybe add a check for itr2 != L2.end() to that.

Answer 4

可以取消引用标准列表迭代器，只要它在列表的范围内即可，即[list.begin（），list.end（）），而列表不为空。

Answer 5

The other answers are correct that a bad list iterator is being dereferenced because of two bugs. Looking at your pastebin,

This condition is backwards:

if ( (*itr1 == *itr2) && (itr2 != L2.end()) ) {

It should be

if ( (itr2 != L2.end()) && (*itr1 == *itr2) ) {

in order to check that itr2 is valid before using it. Also, the first condition

if ( L1.empty() && L2.empty() ) {
            cout << "Returning an empty list because the two arugment lists were empty\n\n";

should be a disjunction:

if ( L1.empty() || L2.empty() ) {
            cout << "Returning an empty list because at least one of the two argument lists was empty\n\n";

but it's not even really necessary.

(Oh, and, are you aware of set_intersection , which is part of the standard library?)