while loop continues after sentinel value is reached - c++11

Question

I'm currently confused on while loops, specifically why the loop still runs everything else in the loop specifically after the sentinel value is reached...
here's the example of my code below:
while (quit != -1) {
cout<<"Options: a)ir; w)ater; s)teel; q)uit \n";
char option=' ';
cin>>option;
option = toupper(option);
while( option != 'A' && option != 'W' && option != 'S' && option != 'Q'){
cin.clear();
cout<<"INVALID INPUT!! \n";
cin>>option;
option = toupper(option);
}
switch(option){
case 'Q':
cout<<"You Have Quit!";
quit = -1;
break;
}
cout<<"Please input the distance: \n";
double distance=0;
cin>>distance;
while(distance < 0){
cin.clear();
cout<<"INVALID INPUT \n";
cin>>distance;
}
switch(option){
case 'A':
seconds = distance/speed_a;
cout<< "Time travled: " << setprecision(2) << fixed << seconds << "\n";
break;
case 'W':
seconds = distance/speed_w;
cout<< "Time travled: " << setprecision(2) << fixed << seconds << "\n";
break;
case 'S':
seconds = distance/speed_s;
cout<< "Time travled: " << setprecision(2) << fixed << seconds << "\n";
break;
}
}
basically when the user enters the 'q' character it runs the quit switch statement... it does quit the loop but it still asks for the distance traveled before ending the loop..
output is shown below:
Options: a)ir; w)ater; s)teel; q)uit
q
You Have Quit!
Please input the distance:
0
--------------------------------
Process exited after 2.944 seconds with return value 0
Press any key to continue . . .
is there any way to edit the code so it immediately quits the loop or is this just the way all loops are pre-programmed?
any advice would be appreciated

Answer 1

The break statement Inside a switch case only exits the switch block, not the outer while loop. Your code continues to run until the next evaluation of the while loop condition.