Question

我试图从数组中提取单个值并将它们放入文本中，但是语法会出现错误。

#include <iostream>
#include <ostream>
using namespace std;
// PV, PM, FUE, RAP, PRE, SAB, ESP
int luch_bas [7] = {6,3,5,3,2,1,2};

int main ()
{ cout << "Tiene los atributos siguientes: \n";
  cout << "Puntos de Vida (PV)... " luch_bas[0] "\n";
  cout << "Puntos de Magia (PM)..." luch_bas[1] "\n";
  cout << "Fuerza (FUE) ..." luch_bas[2] "\n";
  cout << "Rapidez (RAP) ..." luch_bas[3] "\n";
  cout << "Precisión (PRE) ..." luch_bas[4] "\n";
  cout << "Sabiduría (SAB) ..." luch_bas[5] "\n";
  cout << "Espíritu (ESP) ..." luch_bas[6] "\n";
}

我不断得到的错误日志是＆＃34;预期＆＃39 ;;＆＃39;之前＆＃39; luch_bas＆＃39;＆＃34;，但我不确定遗失的地方;应该去？我确信有更好的方法来编写代码;我还在学习。

Answer 1

试试这个

  int main ()
    { cout << "Tiene los atributos siguientes: \n";
      cout << "Puntos de Vida (PV)... \t" << luch_bas[0]<< "\n";
     ...
    }

Answer 2

我一直得到的错误日志是“预期的”;'在'luch_bas'之前，但我不确定遗失的地方;应该去吗？

您不会像错误消息所暗示的那样放置;。您希望分别链接输出操作员调用。

std::ostream& operator<<(std::ostream& os, Type t)会返回std::ostream&引用，如您所见。

正确的方法是链接operator<<()来电：

 cout << "Puntos de Vida (PV)... " << luch_bas[0] << "\n";
                                // ^^             ^^

Answer 3

应该是：

{
    "hooks": {
        "grunt": false
    }
}

Answer 4

使用更多＆＃34;流媒体＆＃34;运营商

#include <iostream>
#include <ostream>
using namespace std;
// PV, PM, FUE, RAP, PRE, SAB, ESP
int luch_bas [7] = {6,3,5,3,2,1,2};

int main ()
{ cout << "Tiene los atributos siguientes: \n";
  cout << "Puntos de Vida (PV)... " << luch_bas[0] << "\n";
  cout << "Puntos de Magia (PM)..." << luch_bas[1] << "\n";
  cout << "Fuerza (FUE) ..." << luch_bas[2] << "\n";
  cout << "Rapidez (RAP) ..." << luch_bas[3] << "\n";
  cout << "Precisión (PRE) ..." << luch_bas[4] << "\n";
  cout << "Sabiduría (SAB) ..." << luch_bas[5] << "\n";
  cout << "Espíritu (ESP) ..." << luch_bas[6] << "\n";
}

如何在文本中打印数组中的单个值？

4 个答案: