C++ OOP 分段错误

Question

我在学院有一项任务。我正在制作一个棋盘，上面有棋子。我有一种step()移动它们的方法。但是我需要打印移动的结果。尝试打印后出现分段错误。并且有没有其他方法可以声明带有符号的棋盘。在我看来，它没有正确声明。(是的，棋子应该像"01""10".稍后我将使用它来为每种作品类型制定规则(。

#include <iostream>
#include <stdio.h>
#include <string>
#define n 8
using namespace std;
int convert(char c) {
if(c=='a')
return 0;
else if(c=='b')
return 1;
else if(c=='c')
return 2;
else if(c=='d')
return 3;
else if(c=='e')
return 4;
else if(c=='f')
return 5;
else if(c=='g')
return 6;
else 
return 7;
}
class chess {
private:
string board[n][n]= {{"02","03","04","06","05","04","03","02"}, //chessboard
{"01","01","01","01","01","01","01","01"}, //black
{"00","00","00","00","00","00","00","00"},
{"00","00","00","00","00","00","00","00"},
{"00","00","00","00","00","00","00","00"}, //00-empty cells
{"00","00","00","00","00","00","00","00"},
{"11","11","11","11","11","11","11","11"}, //white
{"12","13","14","16","15","14","13","12"}};
public:
chess() {}
void print() {
cout<<"  _";  //board marking
for (int i='a'; i<='h'; i++)
printf("_%c_",i);
cout<<endl;
for (int i=0; i<n; i++) { 
printf("%d| ", i+1); //board marking
for (int j=0; j<n; j++) {
printf("%.2s ",board[i][j].c_str()); //print chess pieces
}
cout<<'n';
}
cout<<'n';
}
void step() { //move on another cell
char figc, fign, cellc, celln;
cout<<"Piece, char: "; //choose a chess piece
cin>>figc;
cout<<"Piece, num: ";
cin>>fign;
cout<<"Step, char: "; //move on cell - ...
cin>>cellc;
cout<<"Step, num: ";
cin>>celln;
board[celln-1][convert(celln)]=board[fign-1][convert(fign)];
board[fign-1][convert(fign)]="00";
}
};
int main() {
chess A;
A.step();
A.print();
return 0;
}

例。截图

Answer 1

编译代码

g++ -g -faddress=sanitize main.cpp

显示以下输出：

#0 0x7fdc8e7c4b40 in std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::_M_assign(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const&) (/usr/lib/x86_64-linux-gnu/libstdc++.so.6+0x124b40)
#1 0x7fdc8e7c4f18 in std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::operator=(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const&) (/usr/lib/x86_64-linux-gnu/libstdc++.so.6+0x124f18)
#2 0x56310c7ad092 in chess::step() /z/bob.cpp:65
#3 0x56310c7a9c10 in main /z/bob.cpp:72
#4 0x7fdc8e0b8b96 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x21b96)
#5 0x56310c7a9a19 in _start (/z/a.out+0x1a19)

第 65 行是：

board[fign-1][convert(fign)]="00";

在其中我们看到fign，这是一个用作整数的char。

由于字母和数字的 ascii 范围在 48-122 之间，这超出了数组的大小。

用

board.at(fign-1).at(convert(fign))="00";

会立即告诉你这一点，因为.at()会进行边界检查。

Answer 2

• 你应该fign & celln数据类型作为int.

• 而且您还需要有一些条件，检查输入字符 (a-z(，而且数字也是< n.

  #include <iostream>
  #include <stdio.h>
  #include <string>
  #define n 8
  using namespace std;
  int convert(char c) {
  if(c=='a')
  return 0;
  else if(c=='b')
  return 1;
  else if(c=='c')
  return 2;
  else if(c=='d')
  return 3;
  else if(c=='e')
  return 4;
  else if(c=='f')
  return 5;
  else if(c=='g')
  return 6;
  else 
  return 7;
  }
  class chess {
  private:
  string board[n][n]= {{"02","03","04","06","05","04","03","02"}, //chessboard
  {"01","01","01","01","01","01","01","01"}, //black
  {"00","00","00","00","00","00","00","00"},
  {"00","00","00","00","00","00","00","00"},
  {"00","00","00","00","00","00","00","00"}, //00-empty cells
  {"00","00","00","00","00","00","00","00"},
  {"11","11","11","11","11","11","11","11"}, //white
  {"12","13","14","16","15","14","13","12"}};
  public:
  chess() {}
  void print() {
  cout<<"  _";  //board marking
  for (int i='a'; i<='h'; i++)
  printf("_%c_",i);
  cout<<endl;
  for (int i=0; i<n; i++) { 
  printf("%d| ", i+1); //board marking
  for (int j=0; j<n; j++) {
  printf("%.2s ",board[i][j].c_str()); //print chess pieces
  }
  cout<<'n';
  }
  cout<<'n';
  }
  void step() { //move on another cell
  char cellc, figc;
  int fign,celln; //change datatype as int
  cout<<"Piece, char: "; //choose a chess piece
  cin>>figc;
  cout<<"Piece, num: ";
  cin>>fign;
  cout<<"Step, char: "; //move on cell - ...
  cin>>cellc;
  cout<<"Step, num: ";
  cin>>celln;
  // cout<<"print=>"<<convert(cellc)<<endl;
  // cout<<celln-1<<"  "<<convert(cellc)<<endl;
  // cout<<fign-1<<"  "<<convert(figc)<<endl;
  // cout<<board[celln-1][convert(cellc)].c_str();
  // cout<<board[fign-1][convert(figc)].c_str();
  board[celln-1][convert(cellc)]=board[fign-1][convert(figc)];
  board[fign-1][convert(figc)]="00";
  }
  };
  

Answer 3

你正在使用一个字符fign并在其上做-1。

在这种情况下最好使用 atoi --

board[atoi(&fign)-1][convert(fign)]="00";

输出--

__a__b__c__d__e__f__g__h_
1| 02 03 04 06 05 04 03 02
2| 01 01 01 01 01 01 01 01
3| 00 00 00 00 00 00 00 00
4| 00 00 00 00 00 00 00 00
5| 00 00 00 00 00 00 00 00
6| 00 00 00 00 00 00 00 00
7| 11 11 11 11 11 11 11 00
8| 12 13 14 16 15 14 13 12

为什么你不应该执行字符输入？

好吧，我在那个上面添加了一个 printf %d，给了我这个，这是不言自明的。

.
printf ("nThe result of bad operation : %dn", fign - 1);
.
.
The result of bad operation : 54