switch function

hi my program keeps skiping my switch function any ideas??

Code:

int main (void)
{
  people datab;
  store tdata;
  char any;
  
  char choice;
  char choice2;
  float count = 0;
  float count2 = 0;
  int c; 
  
  tdata.clearary();
  
  cout<<"\nThis program will allow the user to create and then control a database. \n";
  
   
  cout<<"what would you like to do? \n To enter a new record press n \nTo delete a record press d \nTo view a full list of the users with details press f \nTo view serch options press s\n to quit press q :- ";
  cin>>choice;
   

 while(count == 0) 
 {
      choice = getchar ();
	  switch(choice)
	   {
	    case 'n':
		 {    
		      system("cls");
			  tdata.load();
		      tdata.enter(1);
			  tdata.save();
		  count = 1;
		  break;
		 } 
		 
    	case 'd':
		 {
		  //delete a record ??
		  count = 1;
		  break;
		 }
		 
		 case 'f':
		 {
		  system("cls");
		  tdata.list();
		  count = 1;
		  break;
		 }

p.s this is just a snipit also using quincy

Choice will be '\n' the first time in the loop. You're not handling that case.
Kurt

could you explain how

Code:

  cin>>choice;

will read a char and leave '\n' in the buffer.

Code:

choice = getchar ();

getchar will return '\n'.
Kurt

so how do i fix that

I would remove the

Code:

cin>>choice;

Kurt

but wouldnt that stop me getting the users answer

Code:

choice = getchar ();

Should reads the users choice as well.

BTW it is never a good idea to mix c-style with c++ style IO.

Kurt

could you fix this for me as well as that worked fine

Code:

 char k = 'y'; 
	 while (k == "y");

If you use doublequotes, the char inside will be treated as a C-style string. That's wrong, although the compiler might be extra smart about it. Use single quotes for single chars.

Code:

while (k == 'y') {

That shouldn't compile.
try

Code:

	 char k = 'y'; 
	 while (k == 'y'); // this would be an infinite loop. Do you really want the ;

Kurt

thanks i didnt notice that

k i got it mostly working but for some reason the first two questions print in a single line

Code:

 void people::enter(int i) 
   {
    char k = 'y'; 
	 while (k == 'y') 
	   { 
	    cout << "Enter the name of the person ";
	     gets(np[i].name);
        cout << "Enter the phone number of the person ";
	     gets(np[i].phone);
		cout << "Enter the full address of the person ";
		 gets(np[i].address); 
		cout << "Enter the Email address of the person";
		 gets(np[i].email);
		cout << "enter the name of the party that the person belongs to ( conservative, labour, liberal, other)";
		 gets(np[i].party);
		cout << "Do you wish to enter another member ? (y/n)";
		cin  >> k;
		}
	}

any idea

Let me guess. Before calling that function you are using getchar() to read a character.
You will have to get rid of a leftover '\n' in the input buffer.
cin.ignore() could help.
Kurt