Pushing elements to a vector within a ranged based for loop

Question

Could someone explain to me why this simple code:

#include <iostream>
#include <vector>

int main()
{
    std::vector<int> v = {0, 1, 2, 3, 4, 5};

    for(auto i: v)
    {
        std::cout << i << ' ';
    }

    std::cout << std::endl;

    for(auto i: v)
    {
        std::cout << i << ' ';
        v.push_back(4);
    }

    std::cout << std::endl;

    for(auto i: v)
    {
        std::cout << i << ' ';
    }
}

Code posted here: http://cpp.sh/5kxdu

Outputs:

0 1 2 3 4 5 
0 0 2 3 4 5 
0 1 2 3 4 5 4 4 4 4 4 4

While I would expect:

0 1 2 3 4 5 
0 1 2 3 4 5 
0 1 2 3 4 5 4 4 4 4 4 4

I know it's bad to modify the vector during the for loop (was just trying something else when I hit this). After v declaration, v.capacity() returns 6 and obviously, adding v.reserve(12) after v is declared fixes the issue. There is definitely something related to the vector reallocating itself to expand its capacity during the push_back operation.

But I'm wondering if that's just an undetermined behaviour and I was just lucky that only the second element was wrong or if this is something predictable and explainable (considering how the for loop works).

NathanOliver · Accepted Answer · 2016-01-21T20:30:59.547

7

A ranged based for loop has the form of

{
    auto && __range = range_expression ; 
    for (auto __begin = begin_expr,
         __end = end_expr; 
         __begin != __end; ++__begin) { 
        range_declaration = *__begin; 
        loop_statement 
    } 
}

and sets the begin and end iterators at the start of the loop. When you call push_back it always invalidates the end iterator and if size grows larger than the capacity then all iterators are invalidated.

So using push_back is going to cause undefined behavior as you are using iterators that are no longer valid.

edited Jan 21 '16 at 20:30

answered Jan 21 '16 at 20:29

NathanOliver

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So the output could have been much more messed up than it was, right? – jpo38 Jan 21 '16 at 20:30
1

@jpo38 Yes. Since it has UB it could have cause daemons to fly out of the computer. – NathanOliver Jan 21 '16 at 20:31

score 2 · Answer 2 · answered Jan 21 '16 at 20:30

2

After the first push_back the vector reallocates its buffer and iterators in the ranged for loop are invalidated.

answered Jan 21 '16 at 20:30

101010

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I would say it may reallocate buffers. – Slava Jan 21 '16 at 20:40
@Slava Since the OP says the capacity is 6 it is guaranteed that it will reallocate. – NathanOliver Jan 21 '16 at 20:44
@NathanOliver yea missed that part in question, I am very sloppy today :( – Slava Jan 21 '16 at 20:49
@Slava `for( Slava ) give("Coffe");` – NathanOliver Jan 21 '16 at 20:50
@Slava IMHO either the vector reallocate its buffer or it allocates a new one . Plural (i.e., buffers) I think it would be misleading, since a vector has one buffer. – 101010 Jan 21 '16 at 20:51

score 1 · Answer 3 · edited May 23 '17 at 12:04

1

This is because the iterator used by the for loop can be invalidated during the push back. Under the hood the vector uses a buffer, and the push_back can cause the array to be reallocated.

More references here: http://www.cplusplus.com/forum/general/5189/

What happens under the hood of vector::push_back memory wise?

edited May 23 '17 at 12:04

Community

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answered Jan 21 '16 at 20:32

Greg Hilston

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Pushing elements to a vector within a ranged based for loop

3 Answers3