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Help please

omerl1232 (2)
Ive been trying to think of a way to include a yes or no statement at the end of each pattern how ever i am clueless as to how to do it.. any help would be much appreciated :D
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#include <iostream>
using namespace std;

int main(int argc, const char * argv[])
{
    int i, j,k=0,  rows, select,space;
    char symb, choice;
    cout << "***This program prints patterns***"<< endl;
    
    cout <<"Pattern Selection"<< endl;
    cout << "1.Square"<<endl;
    cout <<"2.Right Triangle"<<endl;
    cout<< "3.Pyramid"<< endl;
    
    
    cout << "\nSelect a Pattern (1/2/3)"<< endl;
    cin >> select;
    
    cout << "What symbol to print (@, &, $):" << endl;
    cin >> symb;
    
    switch (select)
    
    {
    
        case 1:
        
            
        cout << "How many rows to print? (maximum 20) " << endl;
        cin >> rows;
            
            for ( int i =1; i <= rows; i++)
            {
                for ( int j = 1; j <=4; ++j)
                    
                    cout << symb;
                    cout << endl;
            }
            

            cout << "Do you wish to continue? " << endl;
            
            cin >> choice;
            
            if (choice == 'n')
                return 0;
            
            else if (choice == 'y')
                
                
        
            
        case 2:
                
                cout << "\nSelect a Pattern (1/2/3)"<< endl;
            cin >> select;
            
            cout << "What symbol to print (@, &, $):" << endl;
            cin >> symb;

            
        
            cout << "Enter Number of rows: " << endl;
            cin>> rows;
            
            for (i =1; i<= rows; ++i)
                
            {
                for ( j =1; j<=i; ++j)
                    
                    cout << symb;
                cout << endl;
            }
        
            cout << "Do you wish to continue? " << endl;
            
            cin >> choice;
            
            if (choice == 'n')
                return 0;
            
            else if (choice == 'y')
        
        
    
            
        case 3:
            {
                
                cout << "\nSelect a Pattern (1/2/3)"<< endl;
                cin >> select;
                
                cout << "What symbol to print (@, &, $):" << endl;
                cin >> symb;

         
                
                
                cout<<"Enter the number of rows: ";
                cin>>rows;
                
                for(i=1;i<=rows;++i)
                {
                    for(space=1;space<=rows-i;++space)
                    {
                        cout<<" ";
                    }
                    while(k!=2*i-1)
                    {
                        cout << symb;
                        ++k;
                    }
                    k=0;
                    cout<<"\n";
                }
                
                cout << "Do you wish to continue? " << endl;
                
                cin >> choice;
                
                if (choice == 'n')
                    
                    cout << "Thank you for using the System." << endl;
                return 0;
                
                
                  if ( choice == 'y')
                      
                      return true;
                

                
                    
            }
            
            
    }
}
TheKingOfTyrants (232)
Your code is asking if the user wants to build another pattern, yes? You'd have to include some type of loop, probably a do-while loop as you want the code to run at least once.
Something like so:
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//explain the program to user
do {
    //ask user for pattern, symbol, and row number preference
    switch ( select ) {
        //strictly output pattern
    }
    //prompt user if they wish to continue
} while ( choice == 'y' );


Also, be careful. Your code has a bit of redundancy and repetition. If I pick symbol 2 or 3, I get the same prompt in the switch statement.
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