Using erase-remove_if idiom

The correct code is:

stopPoints.erase(std::remove_if(stopPoints.begin(),
                                stopPoints.end(),
                                [&](const stopPointPair stopPoint)-> bool 
                                       { return stopPoint.first == 4; }), 
                 stopPoints.end());

You need to remove the range starting from the iterator returned from std::remove_if to the end of the vector, not only a single element.

"Why?"

• std::remove_if swaps elements inside the vector in order to put all elements that do not match the predicate towards the beginning of the container. This means that if the predicate (body of the lambda function) returns true, then that element will be placed at the end of the vector.

• remove_if then **returns an iterator which points to the first element which matches the predicate **. In other words, an iterator to the first element to be removed.

• std::vector::erase erases the range starting from the returned iterator to the end of the vector, such that all elements that match the predicate are removed.

More information: Erase-remove idiom (Wikipedia).

The method std::vector::erase has two overloads:

iterator erase( const_iterator pos );
iterator erase( const_iterator first, const_iterator last );

The first one only remove the element at pos while the second one remove the range [first, last).

Since you forget the last iterator in your call, the first version is chosen by overload resolution, and you only remove the first pair shifted to the end by std::remove_if. You need to do this:

stopPoints.erase(std::remove_if(stopPoints.begin(),
                                stopPoints.end(),
                                [&](const stopPointPair stopPoint)-> bool { 
                                    return stopPoint.first == 4; 
                                }), 
                 stopPoints.end());

The erase-remove idiom works as follow. Let say you have a vector {2, 4, 3, 6, 4} and you want to remove the 4:

std::vector<int> vec{2, 4, 3, 6, 4};
auto it = std::remove(vec.begin(), vec.end(), 4);

Will transform the vector into {2, 3, 6, A, B} by putting the "removed" values at the end (the values A and B at the end are unspecified (as if the value were moved), which is why you got 6 in your example) and return an iterator to A (the first of the "removed" value).

If you do:

vec.erase(it)

...the first overload of std::vector::erase is chosen and you only remove the value at it, which is the A and get {2, 3, 6, B}.

By adding the second argument:

vec.erase(it, vec.end())

...the second overload is chosen, and you erase value between it and vec.end(), so both A and B are erased.