从int*到int(&C)的转换

Question

我一直在尝试编译，并玩了一圈的和数，但仍然不能弄清楚错误是什么。有什么想法吗？

qsort.cc:22:23: error: no matching function for call to ‘qsort<int>::quicksort(std::vector<int, std::allocator<int> >*)’
qsort.cc:22:23: note: candidate is:

qsort.h:16:6: note: void qsort<T>::quicksort(std::vector<T>&) [with T = int]
qsort.h:16:6: note:   no known conversion for argument 1 from ‘std::vector<int, std::allocator<int> >*’ to ‘std::vector<int, std::allocator<int> >&’

标题：

template <class T>
class qsort
{
public:

void quicksort(vector<T> &v);
void qusort(vector<T> &v, int left, int right);
void print(vector<T> &v);
};

template <class T>
void qsort<T>::quicksort(vector<T> &v)
{
qusort(&v, 0, 0);
}

template <class T>
void qsort<T>::print(vector<T> &v)
{
    for(int i = 0; i < v.size(); i++)
    {
    cout << v[i] << endl;
    }
}

int main()
{
qsort<int> asort;
vector<int> v;

v.push_back(2);
v.push_back(1);
v.push_back(7);
v.push_back(3);
v.push_back(8);
v.push_back(4);
v.push_back(0);
v.push_back(9);
v.push_back(5);
v.push_back(6);

asort.quicksort(&v);
asort.print(&v);

return 0;
}

Answer 1

成员函数通过引用获取参数。它们不接受指针（地址运算符&返回的指针）。只需传递对象，引用就会绑定：

asort.quicksort(v);
asort.print(v);