迭代二维STL向量c ++

Question

我正在尝试打印我正在制作的游戏中玩家的动作历史.在每轮结束时,每个玩家都在正方向或负方向上移动了一些数量,并将其记录为运动矢量中的int.最终我想要为每个玩家绘制移动方向与时间的关系,但是我无法从2d向量中提取数据.

所以我尝试的第一件事就是迭代并打印所有元素,但是这不能编译:

void output_movement(const std::vector< std::vector<int> > & movement){

    std::vector< std::vector<int> >::iterator row;
    std::vector<int>::iterator col;
    for (row = movement.begin(); row != movement.end(); ++row) {
         for (col = row->begin(); col != row->end(); ++col) {
            std::cout << **col;
         }
    }

}

编译器提供此错误消息,我不太明白:

hg_competition.cpp:45: error: no match for ‘operator=’ in ‘row = ((const std::vector<std::vector<int, std::allocator<int> >, std::allocator<std::vector<int, std::allocator<int> > > >*)money_movement)->std::vector<_Tp, _Alloc>::begin [with _Tp = std::vector<int, std::allocator<int> >, _Alloc = std::allocator<std::vector<int, std::allocator<int> > >]()’
/usr/include/c++/4.4/bits/stl_iterator.h:669: note: candidates are: __gnu_cxx::__normal_iterator<std::vector<int, std::allocator<int> >*, std::vector<std::vector<int, std::allocator<int> >, std::allocator<std::vector<int, std::allocator<int> > > > >& __gnu_cxx::__normal_iterator<std::vector<int, std::allocator<int> >*, std::vector<std::vector<int, std::allocator<int> >, std::allocator<std::vector<int, std::allocator<int> > > > >::operator=(const __gnu_cxx::__normal_iterator<std::vector<int, std::allocator<int> >*, std::vector<std::vector<int, std::allocator<int> >, std::allocator<std::vector<int, std::allocator<int> > > > >&)

任何帮助是极大的赞赏!

Answer 1

你需要使用const_iteratorif vector是一个const引用.此外,要输出col您只需要取消引用一次.

void output_movement(const std::vector< std::vector<int> > & movement){

    std::vector< std::vector<int> >::const_iterator row;
    std::vector<int>::const_iterator col;
    for (row = movement.begin(); row != movement.end(); ++row) {
         for (col = row->begin(); col != row->end(); ++col) {
            std::cout << *col;
         }
    }
}

编辑:使用typedef将使您的代码更具可读性

typedef std::vector<int> Vector;
typedef std::vector<Vector> DoubleVector;

void output_movement(
    const DoubleVector& movement
)
{
    for (DoubleVector::const_iterator row = movement.begin(); row != movement.end(); ++row) {
         for (Vector::const_iterator col = row->begin(); col != row->end(); ++col) {
            std::cout << *col;
         }
         std::cout << std::endl;
    }
}

Answer 2

vector声明了2D const,因此您需要使用const_iterator而不是iterator.

你也不应该加倍取消引用col.它是一个迭代器,所以你只需要解除引用一次.

void output_movement(const std::vector< std::vector<int> > & movement){ 

    std::vector< std::vector<int> >::const_iterator row; 
    std::vector<int>::const_iterator col; 
    for (row = movement.begin(); row != movement.end(); ++row) { 
         for (col = row->begin(); col != row->end(); ++col) { 
            std::cout << *col; 
         } 
    } 

}