Question

当我尝试输入不良名称（非数字或字母数字）时，我想抛出例外。第一次运行良好，但是当我从catch调用重新进入功能并再次输入坏名时，我得到此错误：抛出'int'实例后终止终止被调用

int main(int argc, char *argv[]) {
    int numberOfWarrior = atoi(argv[1]);
    int numberOfthief = atoi(argv[2]);
    int numberOfnecromancer = atoi(argv[3]);
    int vecorSize = numberOfnecromancer + numberOfthief + numberOfWarrior;
    vector<Hero*> turnOfPlayer;
    //Enter Warrion Players
    if(numberOfWarrior>0) {
        try {
            enterWarrior(0, turnOfPlayer, numberOfWarrior,"warrior");
            }
        catch (int i) {
            cout << "Invalid name of user. You can only choose letters or numbers. Try again please." << endl;
            enterWarrior(i, turnOfPlayer, numberOfWarrior, "warrior");  // my program terminate when i enter to function from here
        }
    }

void enterWarrior(int index, vector<Hero*> v,int numOfWarrior, std::string Type)
{
    std::string nameOfwarrior;
    for(int i=index; i<numOfWarrior; i++)
    {
        cout << "Please insert " << Type << " number " << i+1 << " name:";
        cin >> nameOfwarrior;
        if(!digitCheck(nameOfwarrior))
            throw i;  // in the second time i get the error here 
        if(Type.compare("warrior")==0) {
            Warrior *warr = new Warrior(nameOfwarrior);
            v.push_back(warr);
        }
        if(Type.compare("thief")==0) {
            Thief *thief = new Thief(nameOfwarrior);
            v.push_back(thief);
        }
        if(Type.compare("necromancer")==0) {
            Necromancer *nec = new Necromancer(nameOfwarrior);
            v.push_back(nec);
        }
    }
}

我不知道该如何解决谢谢

Answer 1

如前所述，只是使用循环，例如：

if(numberOfWarrior>0) {
  int firstIndex = 0.

  for (;;) {
    try {
      enterWarrior(firstIndex, turnOfPlayer, numberOfWarrior,"warrior");
      // if we are here that means all is ok, go out of the loop
      break;
    }
    catch (int i) {
      cout << "Invalid name of user. You can only choose letters or numbers. Try again please." << endl;
      firstIndex = i;
    }
  }
}

在void enterWarrior(int index, vector<Hero*> v,int numOfWarrior, std::string Type)中警告，向量是由值给出的，因此 enterWarrior 会修改向量的副本，因此可以通过void enterWarrior（int index， vector＆v，int numOfWarrior，std :: string Type）`

Type.compare("warrior")==0等不是很容易阅读，您可以将它们替换为(Type == "warrior")等，我们使用的是c ++而不是Java

当上一个“ if”为真时，某些“ if”可以为“ else if”，不进行任何比较。如果不是 Type 的类型，最终的 else 似乎也丢失了，无论如何提示问题

抛出'int'实例后终止调用

1 个答案: