Question

问题是写一个A *搜索代码来逃避用C ++代码给出的迷宫。

我的代码能够编译但无法执行。当我在终端中运行它时，出现错误“分段错误（核心转储）”。任何人都可以帮我解决这个问题吗？感谢。

代码如下：

#include <iostream>     
#include <algorithm>  
#include <cmath> 
#include <string>
#include <vector>

using namespace std;

double heu(int a, int b)
{    return sqrt((a-11)*(a-11) + (b-0)*(b-0));}


class step{
  public:
  int x, y;
  string path;
  int pathcost() const {return path.length() - 1;}
  double totalcost() const {return ((double) pathcost() + heu(x,y));}
  bool comp(const step & , const step & )  ;
  };

bool comp( const step & a, const step & b) {return (a.totalcost() < b.totalcost());}

void explore(step currentstep, char Maze[][12],vector<step> allsteps){ 

   step tempstep;
 if (( currentstep.x != 0 ) && (Maze[currentstep.x -1][currentstep.y] != '1'))
   {tempstep.x = currentstep.x -1; tempstep.y = currentstep.y;
   tempstep.path = currentstep.path + 'W';
   allsteps.push_back(tempstep); }

if (( currentstep.x != 11 ) && (Maze[currentstep.x +1][currentstep.y] != '1'))

  { tempstep.x = currentstep.x +1; tempstep.y = currentstep.y;
   tempstep.path = currentstep.path + 'E';
   allsteps.push_back (tempstep);}

if (( currentstep.y != 0 ) && (Maze[currentstep.x][currentstep.y -1] != '1'))
 {tempstep.x = currentstep.x; tempstep.y = currentstep.y -1;
   tempstep.path = currentstep.path + 'N';
   allsteps.push_back (tempstep);
  }

if (( currentstep.y != 11 ) && (Maze[currentstep.x][currentstep.y +1] != '1'))
  {tempstep.x = currentstep.x; tempstep.y = currentstep.y +1;
   tempstep.path = currentstep.path + 'S';
   allsteps.push_back (tempstep);}
}


int main()
{
string txt =     "000100000000000100111111100100000000000000000000011001001101001001001000001000001000001000001000001111101011100000001000000111101000000100000000";


char Maze[12][12];
for (int i=0; i<12; i++){
for( int j =0; j<12;j++){
     int k=i*12+j;
     Maze[j][i]=txt[k];
     }}

vector<step> allsteps;
step currentstep;
currentstep.x=0;currentstep.y=11;
currentstep.path="";
vector<step>::iterator itr;

while ((currentstep.x != 11) && (currentstep.y!=0)){
  explore(currentstep, Maze,allsteps);
  sort(allsteps.begin(), allsteps.end(),comp);
  itr=allsteps.begin();
  currentstep = *itr;}
  cout<<currentstep.path;
 system ("PAUSE");
  }

Answer 1

使用gcc编译代码我得到以下段错误信息：

#0  0x00000000004016e2 in step::operator= (this=0x7fffffffdb60) at main.cpp:13
#1  0x0000000000401468 in main () at main.cpp:73

如果我走这些路线：

currentstep = *itr;}

并且

class step{

也许它可以帮助你找到一些东西。

检查向量的内容，并打印当前的迭代。

只是一个注释

void explore(step currentstep, char Maze[][12],vector<step> allsteps)

应该是

void explore(step currentstep, char Maze[][12],vector<step>& allsteps)

编辑：当您访问

中的第一个元素时，问题是您将参数指定为值而不是引用

 itr=allsteps.begin();

allsteps为空，因此*itr将是一个无效的迭代器，并会对您的代码进行分段。

Answer 2

您是否尝试过设置断点并调试代码？

 cout << "this is a test line, it will print if my program gets past this point." << endl;
 cin.get();

Answer 3

更改您的代码以通过引用而不是按值传递所有步骤：

void explore(step currentstep, char Maze[][12],vector<step>& allsteps){

不

void explore(step currentstep, char Maze[][12],vector<step> allsteps){

EMERGENT A *搜索迷宫C ++

3 个答案: