函数形参为引用迭代器为啥没有用

Question

vector::iterator Find_VEC(vector::iterator &B_iter, vector::iterator &E_iter, int number)
{
while ( (B_iter != E_iter) && (*B_iter != number) )
{
if (*B_iter == number)
{
return B_iter;
}
++B_iter;
}
if (B_iter >= E_iter)
{
return E_iter;
}

/*cout << *B_iter << endl;*/

}

void main()
{
iter = Find_VEC(iter1, iter2, 4);
}
发现返回地址不存在，为什么引用无效？

Answer 1

https://blog.csdn.net/u014135607/article/details/61417947