C++ 崩溃分段错误：11

Question

我一直在开发一个程序来测试一些字符串操作的可能性。它基本上应该读取字符串列表并能够找到字符的邻居以通过字符串作为电路。代码如下：

    #include <iostream>
    #include <string>
    #include <sstream>
    #include <fstream>

    std::string grid[20]={" "};

    std::string get(int string, int member){
      return grid[string].substr(member,1);
    }
    std::string* getNeighbors(int string, int member){
      std::string neighbors[4];
      neighbors[0]=grid[string-1].substr(member,1);//up
      neighbors[1]=grid[string+1].substr(member,1);//down
      neighbors[2]=grid[string].substr(member-1,1);//left
      neighbors[3]=grid[string].substr(member+1,1);//right
      std::string* p=neighbors;
      return p;//Returns up,down,left,right.
    }
    int main(int argc, char** argv){
      grid[1]="@----^---0";
      grid[2]="abcdefghi0";
      grid[3]="jklmnopqr0";//TODO Change to read of txt*/
      std::string* neighbors;
      for(int i=0;grid[1].length()>i;i++){
        neighbors=getNeighbors(2,1);
        if(neighbors[3]=="-" | neighbors[3]=="^"){
          std::string r=get(1,i);
          (r!="0") ? std::cout<<r:0;//Dangerous. TODO Unknown symbol handling
          std::cout<<neighbors[3];
        }
      }
    }

编译良好，但运行时错误“Segmentation fault: 11”。我正在使用一些我不习惯并且可能滥用的主题和技术。任何帮助都会很棒。

Answer 1

std::string neighbors[4]; 是堆栈分配的。当您外出时getNeighbors它会失去作用域。试着把它放在其他地方（甚至是全球性的，只是作为概念证明）。更好的设计应该将其作为参数传递给您的函数。

void getNeighbors(int string, int member, std::vector<std::string>& neighbors){
      ;
      neighbors[0]=grid[string-1].substr(member,1);//up
      neighbors[1]=grid[string+1].substr(member,1);//down
      neighbors[2]=grid[string].substr(member-1,1);//left
      neighbors[3]=grid[string].substr(member+1,1);//right
    }

编辑：

#include <iostream>
    #include <string>
    #include <sstream>
    #include <fstream>

    std::string grid[20]={" "};
    std::string neighbors[4]; //<---------------------------

    std::string get(int string, int member){
      return grid[string].substr(member,1);
    }
    std::string* getNeighbors(int string, int member){
      neighbors[0]=grid[string-1].substr(member,1);//up
      neighbors[1]=grid[string+1].substr(member,1);//down
      neighbors[2]=grid[string].substr(member-1,1);//left
      neighbors[3]=grid[string].substr(member+1,1);//right
      std::string* p=neighbors;
      return p;//Returns up,down,left,right.
    }
    int main(int argc, char** argv){
      grid[1]="@----^---0";
      grid[2]="abcdefghi0";
      grid[3]="jklmnopqr0";//TODO Change to read of txt*/
      std::string* neighbors;
      for(int i=0;grid[1].length()>i;i++){
        neighbors=getNeighbors(2,1);
        if(neighbors[3]=="-" | neighbors[3]=="^"){
          std::string r=get(1,i);
          (r!="0") ? std::cout<<r:"0";//Dangerous. TODO Unknown symbol handling
          std::cout<<neighbors[3];
        }
      }
    }

neighborsnow 是全球性的（我不喜欢这样，但为 POC 做这项工作）。

Answer 2

getNeighbors() 返回一个指向局部变量的指针。