Re: changing vector while processing one of its elements?
Markus Dehmann wrote:
Hi,
I observed a behavior that I didn't expect: I have a vector of sets. I
iterate over the first of these sets, and while I iterate over it, I
add another set at the end of the vector. I thought that shouldn't
affect the first set I am iterating over, but it does, and I get a
segmentation fault.
It does not affect the first set _as a value_. However, it affects the first
set _as an object (region of memory)_ because inserting an element into the
vector can trigger a re-allocation of the vector and then all current
elements are moved around.
See the example program below.
It works fine if I make set0 a *copy* of v[0], instead of a reference
or a pointer, but I would rather not copy it (this routine is called
very often in my program and I don't see why I should make an
expensive copy).
How can I fix this? Thanks!
int main(int argc, char** argv){
std::vector<std::set<int> > v;
std::set<int> s;
s.insert(1);
s.insert(2);
v.push_back(s);
std::set<int>& set0 = v[0]; // using reference because we don't want
to copy to local var (too expensive)
You are using a reference. Insertions into a vector can invalidate all
references, pointers, and iterators into the vector (for the reasons
mentioned above).
for(std::set<int>::const_iterator it = set0.begin(); it !=
set0.end(); ++it){
std::cout << *it << std::endl;
std::set<int> tmp;
tmp.insert(10);
v.push_back(tmp); // will be v[1], so it shouldn't change set0 or
its
iterator?
}
return 0;
}
As output I get:
1
10
1
10
1
10
...
Segmentation fault
Yup, that the undefined behavior from using an invalidated reference.
Your options include:
a) Use std::list instead of std::vector.
b) Instead of inserting the new sets right away, put them on hold.
c) Use std::vector< some_smart_pointer< std::set<int> > >.
Best
Kai-Uwe Bux