why won't this run?

Question

#include <iostream>

using namespace std;

int main()
{
cout << "11+11=";
int test;
cin>>test;
int Q2;
switch(test){
case 23:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 21:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 20:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 19:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 18:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 17:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 16:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 15:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 24:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 25:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 26:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 27:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 28:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 29:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 30:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 31:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 32:
cout <<"wrong\n 11+11=";cin >>test; cout <<"answer is 22\n 12+12=";cin<<Q2;
break;
case 22:
cout <<"right\n 12+12="; cin>>Q2;}

return 0;
}

Answer 1

Insertion operator << for std::cout.
Extraction operator >> for std::cin. You used both << and >> for cin.

Just write 2 cases: 22 and default, if all the others are the same. Don't list 'em all out. Don't repeat code.

Answer 2

thank you for your response, i will try that.