Question

我正在编写一个具有递归函数的代码。但是它执行了两次，实际上我知道它为什么会发生但是我想阻止它请帮助......

int getMatrices()
{
    cout<<"Enter The Number Of Rows Of First Matrix : ";
    cin>>row_a;
    cout<<"Enter The Number Of Column Of First Matrix : ";
    cin>>column_a;
    cout<<"Enter The Number Of Rows Of Second Matrix (Can Not Be Different From Number Of Column Of First): ";
    cin>>row_b;
    cout<<"Enter The Number Of Column Of First Matrix : ";
    cin>>column_b;
    this->checkPhysibility(column_a, row_b);

    cout<<"\nEnter the first matrix : \n";
    fsor(int i=0;i<row_a;i++)
        for(int j=0;j<column_a;j++)
        {
            cout<<"\tA("<<i+1<<", "<<j+1<<") : ";
            cin>>a[i][j];   
        }
    cout<<"\nEnter the second matrix : \n";
    for(int i=0;i<row_b;i++)
        for(int j=0;j<column_b;j++)
        {
            cout<<"\tB("<<i+1<<", "<<j+1<<") : ";
            cin>>b[i][j];   
        }   
}
int checkPhysibility(int column_a, int row_b)
{
    if(column_a!=row_b)
    {
        cout<<"\n\nYou Entered Wrong Input, The Number Of Column Of First Matrix Can Not Be Different From Row Of Second Matrix For Making Their Multiplication Physible. Please Re-Enter The Values.\n\n";
        this->getMatrices();
    }

}

在调用checkPhysibility方法之后，它会检查它，但是如果进入if语句并再次执行getMatrices它会产生一个递归链，并且getMatrices函数中的代码（在checkPhysibility调用之后）执行两次我想要破解。 return语句不起作用....

Answer 1

将checkPhysibility的返回类型更改为bool。
我认为不需要在该函数中编写任何消息。由于getMatrice具有获取数据和写入消息的所有代码，因此在该函数中打印消息更有意义。
如果getMatrices返回checkPhysibility，则将false更改为返回。
当您在此时，请将checkPhysibility更改为checkFeasibility。

bool checkFeasibility(int column_a, int row_b)
{
   return (column_a == row_b);
}

int getMatrices()
{
   cout<<"Enter The Number Of Rows Of First Matrix : ";
   cin>>row_a;
   cout<<"Enter The Number Of Column Of First Matrix : ";
   cin>>column_a;
   cout<<"Enter The Number Of Rows Of Second Matrix (Can Not Be Different From Number Of Column Of First): ";
   cin>>row_b;
   cout<<"Enter The Number Of Column Of First Matrix : ";
   cin>>column_b;
   if ( false == checkFeasibility(column_a, row_b) )
   {
      cout<<"\n\nYou Entered Wrong Input, The Number Of Column Of First Matrix Can Not Be Different From Row Of Second Matrix For Making Their Multiplication Physible. Please Re-Enter The Values.\n\n";
      // ???
      return 0;
   }

   cout<<"\nEnter the first matrix : \n";
   fsor(int i=0;i<row_a;i++)
      for(int j=0;j<column_a;j++)
      {
         cout<<"\tA("<<i+1<<", "<<j+1<<") : ";
         cin>>a[i][j];   
      }
   cout<<"\nEnter the second matrix : \n";
   for(int i=0;i<row_b;i++)
      for(int j=0;j<column_b;j++)
      {
         cout<<"\tB("<<i+1<<", "<<j+1<<") : ";
         cin>>b[i][j];   
      }   

   // ???
   return 1;
}

Answer 2

简短的回答是通过＆＃34;深度＆＃34;参数。每次调用增加1。然后你知道函数在哪个级别。

答案很长，这不是一种使用递归的好方法。如果用户输入错误，请跳回循环以重新输入。不要尝试使用递归进行循环，至少在C / C ++中。一些编程Lisp的人会以不同的方式告诉你。

如何打破函数返回链

2 个答案: