迭代在2维STL向量c ++ [英] Iterating over 2-dimensional STL vector c++

问题描述

我目前正在尝试在我正在开发的游戏中列出玩家的动作历史。在每个回合结束时，每个玩家已经在正或负方向上移动了一些量，并且这被记录为运动向量中的int。最后，我想画每个玩家的移动方向与时间的方向，但我无法从2d向量中提取数据。

I'm currently trying to print out a history of movements for players in a game I am working on. At the end of each round every player has moved some amount in the positive or negative direction and this gets recorded as an int in the movement vector. Eventually I'm wanting to plot the directions moved vs time for each player but I'm having trouble extracting the data out of the 2d vector.

所以第一件事试图只是迭代和打印所有的元素，但是这不编译：

So the first thing I tried was to just iterate and print all the elements, however this doesn't compile:

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;
         }
    }

}

给出此错误消息，我不太理解：

The compiler gives this error message which I don't really understand:

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> > > > >&)

任何帮助都非常感谢！

推荐答案

c $ c> const_iterator 如果向量是一个常量引用。此外，要输出 col ，您只需要解除引用一次。

You need to use a const_iterator if the vector is a const reference. Also, to output col you only need to dereference it once.

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typedefs您的代码更易读。

using typedefs will make your code more readable

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;
    }
}

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