std::list remove_if using state in stack - c++

Question

I want to remove elements from a std::list with linear complexity (going through each element in the list only once). I need to do so depending on the value of a variable in the stack:
int somevalue= 5;
int count=0;
mylist.remove_if(
[](MyStructure* s)
{
if (s->somefield==somevalue)
{
count++;
return true;
}
else
return false;
});
Of course, this doesn't work - somevalue is a variable in the stack.
I've tried using template functions, only to realize (after illegal operation on bound member function expression) you can't really use them in this situation. I know I need to make a closure somehow, so I've read this question, but I can't use C++0x yet and I failed to adapt the other answer for my usecase, as I don't really understand if there's some magic to operator.
Alternatively, is there some way to remove an element from a list given the current position of an iterator (without going through the whole list again to find the element)?

Answer 1

In terms of the lambda expression (a c++11 feature), you can capture the somevalue by value like this: [somevalue](...) {...}

Answer 2

You need to capture the variable in the sample code:
int somevalue= 5;
mylist.remove_if( [somevalue](MyStructure* s){ s->somefield==somevalue });
If no C++11 could be used you need to make the functor yourself:
// For static constant check
template <int CheckValue>
struct Equal {
operator()(const MyStructure* s) { return s->somefield == CheckValue; }
};
mylist.remove_if(Equal<5>);
..or..
// For dynamic constant check as the sample code
struct Equal {
Equal(int check_value) : m_check_value(check_value) {}
operator()(const MyStructure* s) { return s->somefield == m_check_value; }
private:
int m_check_value;
};
mylist.remove_if(Equal(somevalue));

Answer 3

You must capture somevalue in your lamdba expression to use it:
Example (live here) :
struct MyStructure
{
int somefield;
};
int main(int argc, char** argv) {
std::list<MyStructure> my_list = { { 1 }, { 2 }, { 1 }, { 3 }, { 2 }, { 1 } };
int somevalue = 2;
my_list.remove_if( [somevalue](MyStructure s){ return s.somefield == somevalue; });
// ^^^^^^
// Capture
for(auto& s : my_list)
std::cout << s.somefield << " ";
return 0;
}

Answer 4

Iterate on the elements to identify which one you want to remove. Use erase to remove the identified elements and continue to iterate from the returned iterator.
Something like that:
int somevalue=5;
std::list<MyStructure*> myList;
// ...
std::list<MyStructure*>::iterator it=myList.begin();
while(it!=myList.end())
{
if ((*it)->somefield==somevalue)
{
it = myList.erase(it);
}
else
{
++it;
}
}