2D数组显示不需要的字符

Question

我在显示2d数组的内容时遇到问题，我在这里搜索了解决方案，但是由于我不确定自己的问题是什么，所以我不确定我问的是正确的问题。

我正在编写一个程序，允许用户打开一个.txt文件，该文件始终包含30 x 30个字符，所有字符之间都用逗号隔开，并且第一行和第一行分别为0、1、2、3 、、、 9、0、1 ，2,3 ,,, 9和每行的结尾将是一个新行。 该程序应在屏幕上显示.txt文件的内容，但不要使用逗号，然后继续进行操作以允许用户搜索文件中字符的位置。

我认为最好的方法是使用带有分隔符的getline函数来填充2d数组，这似乎已经奏效，但当我显示数组的内容时，最后一个字符被重复了，我不确定如果是由于我填充数组的方式或显示数组的方式引起的。

while(!inputFile.eof())
{
   for (int i = 0; i < SIZE; ++i)           // SIZE is defined 30
     {
       for (int j = 0; j < SIZE; ++j)
         {
           getline(inputFile, line,',');    
           aArray[i][j] = line;             // aArray and line are declared as strings     
           std::cout << aArray[i][j];
         }
     }
           cout << endl;
}

这是我的输入：

0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9 
1,*,!,8,0,;,*,a,b,0,8,0,.,y,Z,c,4,4,8,8,8,8,y,y,y,4,–,6,8,! 
2,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,0,0,0,5,5,J,J,J,J,J,J,J,J,j 
3,9,8,7,0,8,0,8,0,A,c,4,4,*,F,F,6,F,K,J,H,G,5,s,H,U,P,2,2,0 
4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 
5,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8 
6,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,Y,Y,Y,0,0,0,0,0,0,0 
7,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 
8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8 
9,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,Y,Y,Y,0,0,0,0,0,0,0 
0,0,1,1,2,2,D,D,5,5,$,£,!,a,A,a,A,a,A,a,A,a,A,a,A,a,A,a,A,8 
1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 
2,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8 
3,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,Y,Y,Y,0,0,0,0,0,0,0 
4,0,1,1,2,2,D,D,5,5,$,£,!,a,A,a,A,a,A,a,A,a,A,a,A,a,A,a,A,8 
5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 
6,0,8,0,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8 
7,P,0,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,Y,Y,Y,0,0,0,0,0,0,0 
8,0,1,1,2,2,D,D,5,5,$,£,!,a,A,a,A,a,A,a,A,a,A,a,A,a,A,a,A,8 
9,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 
0,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8 
1,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,Y,Y,Y,0,0,0,0,0,0,0 
2,0,1,1,2,2,D,D,5,5,$,£,!,a,A,a,A,a,A,a,A,a,A,a,A,a,A,a,A,8 
3,0,1,1,2,2,D,D,5,5,$,£,!,a,A,a,A,a,A,a,A,a,A,a,A,a,A,a,A,8 
4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 
5,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8 
6,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,Y,Y,Y,0,0,0,0,0,0,0 
7,0,1,1,2,2,D,D,5,5,$,£,!,a,A,a,A,a,A,a,A,a,A,a,A,a,A,a,0,8 
8,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,P,Y,Y,Y,0,0,0,0,0,8,0 
9,0,1,1,2,2,D,D,5,5,$,£,!,a,A,a,A,a,A,a,A,a,A,a,A,a,A,a,0,8

这是我的输出：

012345678901234567890123456789 
1*!80;*ab080.yZc448888yyy4–68! 
200000000001112200055JJJJJJJJj 
398708080Ac44*FF6FKJHG5sHUP220 
400000000000000000000000000000 
588888888888888888888888888888 
6PPPPPPPPPPPPPPPPPPPYYY0000000 
700000000000000000000000000000 
888888888888888888888888888888 
9PPPPPPPPPPPPPPPPPPPYYY0000000 
001122DD55$£!aAaAaAaAaAaAaAaA8 
100000000000000000000000000000 
288888888888888888888888888888 
3PPPPPPPPPPPPPPPPPPPYYY0000000 
401122DD55$£!aAaAaAaAaAaAaAaA8 
500000000000000000000000000000 
608088888888888888888888888888 
7P0PPPPPPPPPPPPPPPPPYYY0000000 
801122DD55$£!aAaAaAaAaAaAaAaA8 
900000000000000000000000000000 
088888888888888888888888888888 
1PPPPPPPPPPPPPPPPPPPYYY0000000 
201122DD55$£!aAaAaAaAaAaAaAaA8 
301122DD55$£!aAaAaAaAaAaAaAaA8 
400000000000000000000000000000 
588888888888888888888888888888 
6PPPPPPPPPPPPPPPPPPPYYY0000000 
701122DD55$£!aAaAaAaAaAaAaAa08 
8PPPPPPPPPPPPPPPPPPPYYY0000080 
901122DD55$£!aAaAaAaAaAaAaAa0888888888888888888888888888888

如您所见，最后一个字符被重复了，这不是我想要的。 我是C ++的新手，因此不确定是否使用最佳方法，我花了将近2天的时间来解决我因误将cout << endl而创建的问题。 在我的for循环中，因此如果堆栈溢出，我会联系一些优秀的成员，以获取一些急需的指导。

Answer 1

要扩展@WhozCraig的注释，您不应在循环条件内使用iostream::eof ，因为在读取文件末尾之前，它不会返回true。 相反，这样的事情就足够了（注意有很多方法可以做到这一点）：

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

#define SIZE 30

int main() {

    std::string line;
    std::string aArray[SIZE][SIZE];
    std::ifstream inputFile("file2.txt");

    for (int i = 0; getline(inputFile, line) && i < SIZE; ++i) {

        std::istringstream lineStream(line);
        std::string token;

        for (int j = 0; getline(lineStream, token, ',') && j < SIZE; ++j) {
            aArray[i][j] = token;
            std::cout << aArray[i][j];
        }

        std::cout << std::endl;
    }
}

这将提供所需的输出：

012345678901234567890123456789
1*!80;*ab080.yZc448888yyy4–68!
200000000001112200055JJJJJJJJj
398708080Ac44*FF6FKJHG5sHUP220
400000000000000000000000000000
588888888888888888888888888888
6PPPPPPPPPPPPPPPPPPPYYY0000000
700000000000000000000000000000
888888888888888888888888888888
9PPPPPPPPPPPPPPPPPPPYYY0000000
001122DD55$£!aAaAaAaAaAaAaAaA8
100000000000000000000000000000
288888888888888888888888888888
3PPPPPPPPPPPPPPPPPPPYYY0000000
401122DD55$£!aAaAaAaAaAaAaAaA8
500000000000000000000000000000
608088888888888888888888888888
7P0PPPPPPPPPPPPPPPPPYYY0000000
801122DD55$£!aAaAaAaAaAaAaAaA8
900000000000000000000000000000
088888888888888888888888888888
1PPPPPPPPPPPPPPPPPPPYYY0000000
201122DD55$£!aAaAaAaAaAaAaAaA8
301122DD55$£!aAaAaAaAaAaAaAaA8
400000000000000000000000000000
588888888888888888888888888888
6PPPPPPPPPPPPPPPPPPPYYY0000000
701122DD55$£!aAaAaAaAaAaAaAa08
8PPPPPPPPPPPPPPPPPPPYYY0000080
901122DD55$£!aAaAaAaAaAaAaAa08