what is wrong with this code?printing the odd number

Question

//it is giving 8 as output which is even

#include <iostream>
#include <set>
using namespace std;

int main()
{
// set declaration
set<int> myset{ 1, 2, 3, 4, 5, 6, 7, 8, 9 };

// checking for even elements and removing them
for (auto i = myset.begin(); i != myset.end(); ++i) {
if (*i % 2 == 0) {
myset.erase(i);
i--;
}
}

// Printing the set
for (auto it = myset.begin(); it != myset.end(); ++it)
cout << ' ' << *it;
return 0;
}

Answer 1

The problem on my side appears to be the i--; statement after myset.erase(i).

Removing i-- gives me a correct output of

1 3 5 7 9

while the original code gives me no output at all.

Answer 2

i-- while i was invalidated by erase.
do i=myset.erase();

The second error due to 1%2!=0; So, you erase the first item and then do i--. The right way is to not move iterator in the for statement.
for(auto i=m.begin(); i!=m.end();)
if(*i%2==0) i=m.erase(i);
else ++i;

Answer 3

no, it's iterator on element after. It is the reason why ++i under else branch.

Answer 4

try this input:{1,14 ,12 ,122 ,123}

Answer 5

it means erase return an iterator after of erasing element?

Answer 6

ok thanks