std::erase in C++

Question

I am confused about function of std::erase in C++.

The following code gets the same output before and after std::erase is called.But if iterate through the list after performing std::erase then in output the erased value is not showing. Help me understand std::erase.

#include<bits/stdc++.h>
using namespace std;

int main()
{
    list<int> v;
    v.push_back(12);
    v.push_back(10);
    v.push_back(20);

    list<int>::iterator it;
    it = v.begin();
    printf("%u %d\n", it, *it);
    v.erase(it);
    printf("%u %d\n", it, *it);

    for(it= v.begin(); it!= v.end(); it++)
        cout<< *it<<" ";
    return 0;
}

Output:

"Memory address" 12  
"Memory Address" 12  
10 20

What's actually unclear from the [docs](http://en.cppreference.com/w/cpp/container/list/erase)? Besides that your output doesn't render as you describe. — πάντα ῥεῖ, May 27 '17 at 12:16
Also, [don't include ``](http://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h). — Some programmer dude, May 27 '17 at 12:20

score 9 · Accepted Answer · answered May 27 '17 at 12:15

9

erase invalidates the iterator you give it (plus potentially other ones, depending on the container).
Using and dereferencing this iterator afterwards is undefined behaviour.

answered May 27 '17 at 12:15

Quentin

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score 2 · Answer 2 · answered May 27 '17 at 12:25

According to the C++ Standard relative to the class template list

Effects: Invalidates only the iterators and references to the erased elements.

Thus the program has undefined behavior.

After this statement

v.erase(it);

the iterator it that initially was set as

it = v.begin();

now does not correspond to v.begin() and the output of the loop

for(it= v.begin(); it!= v.end(); it++)
    ^^^^^^^^^^^^^         
    cout<< *it<<" ";

confirms that.

Instead of

v.erase(it);

you could write

it = v.erase(it);

and in this case the returned iterator will indeed correspond to the iterator returned by v.begin()

std::erase in C++

2 Answers2