What's wrong with cin >> num1, num2?

Question

#include 
using namespace std;

int main() {
  char choice;
  int solution, num1, num2;

  cout << \"Menu\";
  cout << \"\\n========\";
  cou         


        

Answer 1

cin >> num1, num2;

This syntax is not correct for what you want. To chain, use

cin >> num1 >> num2;

If you compile with warnings you'll get notified about this by the compiler

int a{}, b{};
std::cin >> a, b;

gives the error:

warning: right operand of comma operator has no effect [-Wunused-value]
std::cin >> a, b;

The whole statement is parsed as

((std::cin >> a), b);

Which consists of two comma-separated expressions. The b in that case doesn't have any effect. If you print the variables after the std::cin line above, you will always get 0 for b

Answer 2

int solution, num1, num2;

This leaves all variables uninitialised. Trying to read from them without a previous assignment is undefined behaviour.

cin >> num1, num2;

The supposed intention of this line is to read into both variables. What happens really is an application of the comma operator, with operands cin >> num1 on the left side and num2 on the right side.

The left side is evaluated and a value is written to num1; the second operand has no effect and leaves num2 in its uninitialised status.

It's as if you had written cin >> num1;.

solution = num1 + num2;

The aforementioned undefined behaviour happens, rendering your entire program invalid.

You can fix the problem as follows:

cin >> num1;
cin >> num2;

Answer 3

You are using cin wrong. Do this instead :

cin >> num1 >> num2;

(And it is always nice to let the user know what to type with one cout or two :)).

You could do cin only once just after cin >> choice, and before the switch. It will save you some lines and will respect the credo :

Do not repeat yourself.

You don't check for divide by zero. If the user try to do num1 / 0, your program will crash.

The return 0 in case 'X' is not necessary and your break won't be reached, so you could remove the return and you will meet the main's return 0 instead, keeping the switch nice and clean.

In case '/' you cast num2 to double. I don't see why you print the result of a int casted to a double. My guess is you wanted the result to be a double. In that case, you would do :

solution = num1 / (double)num2; 

and change solution to be a double.

Lastly, I advise you always initialize your variables since you could face the case where you use unitialized variables and get undefined behavior.