c++ - 从 stringstream 中提取到 2D vector 中

Question

这可能是一个愚蠢的问题，但无论如何:我想从一个 .txt 文件中获取数字，其中包含图形的邻接矩阵，文件的第一行仅包含节点数。

10

-1 5 3 -1 -1 -1 -1 -1 -1 -1

5 -1 -1 4 -1 -1 -1 -1 -1 -1

3 -1 -1 -1 -1 9 7 6 -1 -1

-1 4 -1 -1 2 -1 -1 -1 -1 -1

-1 -1 -1 2 -1 -1 -1 -1 -1 -1

-1 -1 9 -1 -1 -1 -1 -1 -1 -1

-1 -1 7 -1 -1 -1 -1 -1 4 2

-1 -1 6 -1 -1 -1 -1 -1 -1 -1

-1 -1 -1 -1 -1 -1 4 -1 -1 -1

-1 -1 -1 -1 -1 -1 2 -1 -1 -1

为什么它不能以正确的方式工作？输出如下:

10

10 0 0 0 0 0 0 0 0 0

5 0 0 0 0 0 0 0 0 0

3 0 0 0 0 0 0 0 0 0

-1 0 0 0 0 0 0 0 0 0

-1 0 0 0 0 0 0 0 0 0

-1 0 0 0 0 0 0 0 0 0

-1 0 0 0 0 0 0 0 0 0

-1 0 0 0 0 0 0 0 0 0

-1 0 0 0 0 0 0 0 0 0

-1 0 0 0 0 0 0 0 0 0

void beolvas (vector<vector<double> > & mygraph, string filename)
{
    ifstream input(filename);
    stringstream ss;
    char node[30];
char graph_size[2];

while(input.good()){

input.getline(graph_size,'\n');
cout << graph_size << endl;
ss << graph_size;
int graph_sizeINT = atoi(graph_size);
mygraph.resize(graph_sizeINT, vector<double>(graph_sizeINT,-1));
ss.clear();

for(int i=0; i<graph_sizeINT; i++)
    {

          input.getline(node,35,'\n');
          //cout << node << endl;
          ss << node;
         for(int j= 0; j<graph_sizeINT; j++)
          {
            ss.getline(node,' ');
            //cout << node << " ";
            mygraph[i][j] = atof(node);
            cout << mygraph[i][j] << " ";
          }
          cout << endl;
          ss << "";
          ss.clear();
   }
} input.close();   }

感谢您的建议!

Answer 1

您正在使用 getline 和 stringstream，它们是很好的工具，但不是这项工作的正确工具；他们太强大了，需要太多照顾。与其确切地分析它们是如何出错的，不如看看当我们放弃它们以支持流输入时会发生什么:

void beolvas (vector<vector<double> > & mygraph, string filename)
{
  ifstream input(filename.c_str());

  int graph_sizeINT;
  while(input >> graph_sizeINT)
    {
      cout << graph_sizeINT << endl;

      mygraph.resize(graph_sizeINT, vector<double>(graph_sizeINT,-1));

      for(int i=0; i<graph_sizeINT; i++)
        {
          char node[30];
          for(int j= 0; j<graph_sizeINT; j++)
            {
              input >> mygraph[i][j];
              cout << mygraph[i][j] << " ";
            }
          cout << endl;
        }
    }
  input.close();
}