I need to go through a set and remove elements that meet a predefined criteria.
This is the test code I wrote:
#include <set>
#include <algorithm>
void printElement(int value) {
std::cout << value << " ";
}
int main() {
int initNum[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
std::set<int> numbers(initNum, initNum + 10);
// print '0 1 2 3 4 5 6 7 8 9'
std::for_each(numbers.begin(), numbers.end(), printElement);
std::set<int>::iterator it = numbers.begin();
// iterate through the set and erase all even numbers
for (; it != numbers.end(); ++it) {
int n = *it;
if (n % 2 == 0) {
// wouldn't invalidate the iterator?
numbers.erase(it);
}
}
// print '1 3 5 7 9'
std::for_each(numbers.begin(), numbers.end(), printElement);
return 0;
}
At first, I thought that erasing an element from the set while iterating through it would invalidate the iterator, and the increment at the for loop would have undefined behavior. Even though, I executed this test code and all went well, and I can't explain why.
My question:
Is this the defined behavior for std sets or is this implementation specific? I am using gcc 4.3.3 on ubuntu 10.04 (32-bit version), by the way.
Thanks!
Proposed solution:
Is this a correct way to iterate and erase elements from the set?
while(it != numbers.end()) {
int n = *it;
if (n % 2 == 0) {
// post-increment operator returns a copy, then increment
numbers.erase(it++);
} else {
// pre-increment operator increments, then return
++it;
}
}
Edit: PREFERED SOLUTION
I came around a solution that seems more elegant to me, even though it does exactly the same.
while(it != numbers.end()) {
// copy the current iterator then increment it
std::set<int>::iterator current = it++;
int n = *current;
if (n % 2 == 0) {
// don't invalidate iterator it, because it is already
// pointing to the next element
numbers.erase(current);
}
}
If there are several test conditions inside the while, each one of them must increment the iterator. I like this code better because the iterator is incremented only in one place, making the code less error-prone and more readable.
Question&Answers:
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