打印2D std :: string数组

Question

我试图用c ++初始化一个tic-tac-toe板，但输出总是给我六进制值。 有没有办法将它们转换为实际的字符串值？

#include <iostream>
#include<string>

using namespace std;

int main()
{
    string tab[5][5] = { "1","|","2","|","3",
                         "-","+","-","+","-",
                         "4","|","5","|","6",
                         "-","+","-","+","-",
                         "7","|","8","|","9" };

    for(int i = 0; i <= 24; i++)
    {
        cout << tab[i] << endl;
    }
}

Answer 1

你将tab[i]的值发送到cout ，所以你得到了内存地址。

您可能希望更深层次地嵌套项目，例如tab[i][j] 。

Answer 2

tab[i]是一个std::string[] - 也就是std::string的数组，而不是std::string本身。

使用ranged- for ，而不是你的输出。 与标准库容器一样，它与内置数组一起使用：

for (const auto &row : tab) {
  for (const auto &c : row) {
    cout << c;
  }
  cout << endl;
}

Answer 3

当然，在输出位置提出循环的答案是正确的。 如果您碰巧要在应用程序的许多不同位置输出您的tic tac toe字段，您可能更愿意将其封装起来。 一个可能的解决方案是有一个tic tac toe类：

struct TicTacToe : public std::array<std::array<int, 3>, 3> {
    TicTacToe() :
        // not sure, if the initialization will work like that
        std::array<std::array<int, 3>, 3>{{0,0,0},{0,0,0},{0,0,0}}
    {};
};

然后为它定义一个输出运算符：

auto operator << (std::ostream& out, const TicTacToe& field)
    -> std::ostream& {
  return
      out << std::accumulate(std::begin(field),
                             std::end(field),
                             std::string{},
                             [](const std::string& a,
                                const std::array<int, 3> b)
                                 -> std::string {
                               return 
                                   std::accumulate(std::begin(b),
                                                   std::end(b),
                                                   std::string{},
                                                   [](const std::string& a, int b)
                                                       -> std::string {
                                                     return std::string{b < 0 ?
                                                                          "O" :
                                                                          (b > 0 ? "X" : " ")} +
                                                            (a.empty() ? "" : "|")
                                                   }) +
                                   (a.empty() ? "" : "\n-+-+-\n");
                             });

}

请注意，我还没有测试过这段代码。 它旨在为您提供一个想法，而不是复制粘贴的来源。

Answer 4

这更像是非常感谢！

#include <iostream>
#include<string>
using namespace std;

int main()
{
    string tab[5][5] = { "1","|","2","|","3",
                         "-","+","-","+","-",
                         "4","|","5","|","6",
                         "-","+","-","+","-",
                         "7","|","8","|","9" };

    for(int i = 0; i < 5; i++)
    {
        for (int j = 0; j < 5; j++)
        {
            cout << tab[i][j];

            if (j == 4)
            cout << endl;
        }   
    }
}