std::transform behavior when iterating past container.end()

Question

Code:

static int counter = 0;

int add(int x) {
    counter++;
    return ++x;
}

int main() {
    vector<int> b;
    b.push_back(1);
    b.push_back(1);
    b.push_back(1);

    transform(b.begin(),b.end(),b.begin()+2,add);

    for (vector<int>::iterator it = b.begin(); it != b.end(); it++)
        cout << (*it) << endl;
    cout << "counter: " << counter << endl;
}

For me this compiles with no warnings and prints out:

1
1
2
counter: 3

What is happening here in the transform function? How is it that add(...) gets called 3 times but b.end() is not overwritten? Is this undefined behavior?

Steve Jessop · Accepted Answer · 2011-06-05 22:37:42Z

1

Yes, it's undefined behavior. It's your responsibility to avoid running past the end of the vector, not the compiler's.

What do you mean by, "b.end() is not overwritten"? If you mean that you expected the vector to change length, then no, it didn't, you can't change a vector's length this way.

answered Jun 5, 2011 at 22:37

Steve Jessop

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9 Comments

Jeff Chen Over a year ago

by "b.end() is not overwritten" I mean the for-loop that prints out the values in the vector has a termination condition it != b.end(). Since 3 values are printed out I'm assuming b.end() was not overwritten.

Steve Jessop Over a year ago

What would you expect it to do if b.end() "was overwritten"? The phrase means nothing to me. The reason 3 values are printed is that the size of the vector is 3 (although of course that's not guaranteed - your undefined behavior could have broken the program beyond all recognition. On this occasion, though, the behavior observed is consistent with the vector surviving the experience).

Steve Jessop Over a year ago

Oh, are you working from a mental model that vectors are nul-terminated, like C-style strings? They aren't.

Jeff Chen Over a year ago

well i'm assuming transform(iter1,iter2,iter3,func) applies func to the elements in [iter1,iter2) and stores the results starting at iter3. Since iter3 in my code is b.begin()+2 and there are only 3 elements in b, I figured it was a possibility that b.end() would get overwritten (or at least throw an exception when b.end() gets dereferenced).

Steve Jessop Over a year ago

Sorry, I still don't understand what you mean by the difference between b.end() "being overwritten" and "not being overwritten", or how you think you can tell whether or not it has happened. Dereferencing an end iterator is undefined behavior, as far as the language is concerned there's nothing there to overwrite. Some C++ implementations provide debugging versions of their containers that help you detect such errors, but apparently no such mode is active in yours.

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zienkikk · Accepted Answer · 2011-06-05 22:54:50Z

1

The first two parameters to transform specify the range to transform. Since your vector contains 3 elements, std::transform calls add() 3 times. The third parameter is the beginning of the output. You're telling std::transform to start storing your result in the 3rd (index 2) position of your vector. Since std::transform uses operator*() to store the result, you're essentially dereferencing memory you don't own. That leads to undefined behavior. On some platforms, this will crash, on others it will seem to work. On all standard-compliant platforms, however, you will only change the elements that actually exist in the vector.

answered Jun 5, 2011 at 22:54

zienkikk

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2 Comments

Steve Jessop Over a year ago

"On all standard-compliant platforms, however, you will only change the elements that actually exist in the vector." - what do you mean? The standard places no restrictions of any kind on what implementations do with this program. A conforming implementation may do anything, not just change elements in the vector.

zienkikk Over a year ago

I meant that the user will not be able to legally see past the last element of the vector even though transform attempts to write to it.

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