Problems with understanding the switch construction, why is default not executed in the next iteration without break?

Question

C++ / C#

The next iteration default does not work, but it works with break. I use 2*2.
In general, I’ll formulate the question even more simply: if there is a break, the default is performed in the next iteration, without a break, the default is not performed in the next iteration. How does a brick change whether the code is executed or not? It just breaks the code. Why does it affect behavior since I'm in an infinite while(true) loop.

double term() {
  double left = primary();
  Token t = ts.get(); // get the next token from token stream

    while (true) {
    switch (t.kind) {
    case '*':
    cout << "I'm inside! '*'\n";
      left *= primary();
      t = ts.get();
     // break был здесь, но без брика не работает, но ведь дефолт в некст итерации все равно выполняется
    // а если так - то какая разница, есть break или нет, такие дела, как понимать?
    case '/': {
      double d = primary();
      if (d == 0)
        error("divide by zero");
      left /= d;
      t = ts.get();
      break;
    }
    default:
      cout << "I'm inside! 'default'\n";  // (Я должен был быть тут, абсурд?)
      ts.putback(t); // put t back into the token stream
      return left;
    }
  }
}

In general, I did not know that the following case if there was no break are executed without checking. I'm sorry!