Question

在新对象上使用vector.push_back时，我得到nullReferenceException。

在代码段中，您看到我将矢量对象作为指针，但我最初将它作为非指针，我在绝望中更改了它。

我完全逐步完成了BasicSolver的实例化，以确保其中没有问题，并将实例化与push_back分开，以帮助显示push_back是问题发生的地方。

InformedSolver1＆amp; 2是BasicSolver的子级。 vector包含在BasicSolver中，因此也包含在puzzleSolver中。

#include "stdafx.h"
#include "puzzleSolver.h"


PuzzleSolver::PuzzleSolver()
{
   solvers = new vector<BasicSolver*>();
}

void PuzzleSolver::createPuzzle(string input)
{                                             //hitting step-over
   BasicSolver* temp = new BasicSolver(input);// no errors
   solvers->push_back(temp);                  // nullReferenceException
   solvers->push_back(new InformedSolver1(input));
   solvers->push_back(new InformedSolver2(input));
}

这应该是所有相关信息。如果您对导致此问题/如何解决问题有任何想法，请告诉我们！感谢。

编辑：通过评论请求添加了BasicSolver构造函数和依赖项方法还有一点背景：这是一个AI类的数独求解器

BasicSolver::BasicSolver(string input)
{
   peers = new vector<Peer*>();
   squares = new vector<Square*>();
   unsolved = new vector<Square*>();
   solved = new vector<Square*>();
   createStructure(input);
   original = input;
   mistakes = 0;
}

void BasicSolver::createStructure(string input)
{
   try
   {
       createEmptyStructure(peers, squares, unsolved);
      //Parse the puzzle and assign input to squares
      int numCharsToRead = MAX_SQUARES;   //makes sure vector isn't outside its range
      if (input.length() < MAX_SQUARES)   //makes sure string isn't outside its range
         numCharsToRead = input.length(); 
      for (int i = 0; i < numCharsToRead; i++)
      {
         if(input[i] != '.')
            insertValue(input[i], (*squares)[i], unsolved);
      }
   }
   catch(exception e)
   {
      throw e;
   }
}

void BasicSolver::createEmptyStructure(vector<Peer*> *workingPeers, vector<Square*> *workingSquares, vector<Square*> *workingUnsolved)
{
    for (int i = 0; i < MAX_PEERS; i++)
    {
        workingPeers->push_back(new Peer());
    }
   for (int i = 0; i < 81; i++)
   {
      try
      {
          workingSquares->push_back(new Square('.'));

         //Adding the square to its corresponding peers
         (*workingPeers)[i / MAX_ROWS]->addSquare((*workingSquares)[i]); //adds the square into its appropriate row of peers
         (*workingPeers)[(i % MAX_ROWS) + COL_OFFSET]->addSquare((*workingSquares)[i]); //adds the square into its appropriate column of peers
         int tempBoxCol = (i % MAX_ROWS) / MAX_BOX_COLS; //returns the box column (0,1,2)

         if ((i / MAX_ROWS) < BOX_ROW_WIDTH) //if its box is in the first row
         {
             (*workingPeers)[tempBoxCol + BOX_OFFSET]->addSquare((*workingSquares)[i]);
         }
         else if ((i / MAX_ROWS) < (2 * BOX_ROW_WIDTH)) //if its box is in the second row
         {
             (*workingPeers)[BOX_ROW_WIDTH + tempBoxCol + BOX_OFFSET]->addSquare((*workingSquares)[i]);
         }
         else //if the box is in the third row
         {
             (*workingPeers)[2 * BOX_ROW_WIDTH + tempBoxCol + BOX_OFFSET]->addSquare((*workingSquares)[i]);
         }
      }
      catch(exception e)
      {
         throw e;
      }
   }
   *workingUnsolved = *workingSquares;
}

EDIT2：个人测试添加以下行：

vector<BasicSolver*>* test = new vector<BasicSolver*>();
test->push_back(temp);

前

solvers->push_back(temp);

并且它们执行正常，我还注意到在运行时解算器被列为范围之外，即使它是BasicSolver的受保护变量。

Answer 1

问题是调用源没有正确实例化PuzzleSolver。感谢@TheDark建议我检查一下！

push_back上的NullReferenceException

1 个答案: