c ++帮助将向量索引传递给函数

Question

I know how to pass a vector to a function, but how do I pass a vector index to a function, or at least specify which index the function is modifying. For example, I'm working on a Car class and it has a vector if wheel pointers and in order to remove one of the wheels my function looks like this:

Wheel& remove() {
    for (int i = 0; i < wheels.size(); i++) {
        if (wheels[i].position == wheels.at(i)) {
            ??
        }

what do I need to pass to the function in order to specify which wheel I want removed? When a wheel is removed, the position where it was is still there and can be filled by another wheel. Let's say for example the car had 4 wheels...if I wanted to remove the 2nd index in the wheel vector, what does the function argument for remove() need to take in order to do it? Should I pass in the vector and then the specific index....and if so, what does the syntax look like?

Answer 1

You can just pass an integer to specify which one you would like to remove

void RemoveWheel(int i)
{
    if( (i<wheels.size()) and (i>=0) )
        wheels.erase(wheels.begin()+i);
}

http://www.cplusplus.com/reference/stl/vector/erase/

If you want to leave the space for another wheel than you should define wheels as a vector of pointers and just delete object at the i-th position and save NULL instead of it.

vector<Wheel *> wheels;

void RemoveWheel(int i)
{
    if( (i<wheels.size()) and (i>=0) ) {
        delete wheels[i];
        wheels[i] = 0;
    }
}

Answer 2

Your question is not entirely clear to me, but to remove an element from a vector, given an index i, you can do this:

wheels.erase(wheels.begin() + i);

But this would be better:

auto e = std::remove_if(wheels.begin(), wheels.end(),
             [](const Wheel & wheel) {
                 return wheel.position == wheel;
             });
wheels.erase(e,wheels.end());

Although I'm not sure if you want to remove every element that fits that criteria, or just the first. If you would show the logic in pseudocode of what you want to do, that would help.