Removing Characters from String until it becomes empty - c++

Question

I can't get my head around this. I'm trying to remove all occurrences of a certain character within a string until the string becomes empty. I know we can remove all character occurrences from an std::string by using the combination of string::erase and std::remove like so:
s.erase(remove(s.begin(), s.end(), '.'), s.end());
where the '.' is the actual character to be removed. It even works if I try to remove certain characters. Now let's consider the following string: 'abababababababa'. What I'm trying to achieve is to reduce this string to ashes be removing all 'a's for startes, which will leave me with a couple of 'b's. Then remove all those 'b's which will leave me with an empty string. Of course this is just a part of my task but I could narrow it down for this problem. Here's my naive approach based on the upper combination of functions:
string s = "abababababababa";
while (!s.empty()) {
...
s.erase(remove(s.begin(), s.end(), s[0]), s.end());
...
}
Of course it doesn't work, I just can't seem to find out why. By debugging the application I can see how the "s" string is being modified. While the s.erase... works perfectly if I set a character constant for remove's third parameter it fails if I try to use char variables. Here's what the s string looks like after each iteration:
Removing[a] from [abababababababa] Result: baaaaaaa
Removing[b] from [baaaaaaa] Result: a
Removing[a] from [a] Result: -
While I expected 2 operations until a string should become empty - which works, if I hardcode the letters by hand and use s.erase twice - it actually takes 3 iteration. The most frustrating part however is the fact that, while I'm removing 'a' in the first iteration only the first 'a' is removed and all other 'b'.
Why is this happening? Is it the cause of how erase / remove works internally?

Answer 1

You have undefined behavior.
You get the results you get because std::remove takes the value to remove by reference, once s[0] has been removed, what happens to the reference to it then?
The simple solution is to create a temporary variable, assign e.g. s[0] to it, and pass the variable instead.

Answer 2

The behavior of function remove() template is equivalent to:
template <class ForwardIterator, class T>
ForwardIterator remove (ForwardIterator first, ForwardIterator last, const T& val)
{
ForwardIterator result = first;
while (first!=last) {
if (!(*first == val)) {
*result = move(*first);
++result;
}
++first;
}
return result;
}
As you see, the function will move the element different with val to the front of the range.
so in your case "ababababab",
if you call remove() like you did, the original s[0] is 'a', but it will be instead by 'b' during the remove(), the remaining code will remove the 'b', so the result is not right.
Like Joachim say, assign s[0] to a temporary variable.
the code is reference from http://www.cplusplus.com/reference/algorithm/remove/?kw=remove