When I write a digit to a variable, the digit is stored until such time as I do not record another value, but I need to make the old value remain, the type is shifted to the high order, and the new value is written to the youngest ....... example : I have the number 1 in the variable, and when I add the number 2 there, do I need to have 12?

Closed due to the fact that it is necessary to reformulate the question so that it was possible to give an objectively correct answer by the participants αλεχολυτ , user194374, Bald , Denis , Kromster 20 Dec '16 at 13:29 .

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  • one
    There are no bits here ... - Qwertiy ♦
  • How do you get from, say, 34 number 345? This is 34 * 10 + 5. And from 15 - 157? This is 15 * 10 + 7. Well, think about it ... Consider only the sign for signed numbers, if you transmit 5, not -5 ... - Harry
  • Read the book Peter Boole Boolean Algebra , study the operators & , | , && , || , << , >> - nick_n_a

2 answers 2

these are not bitwise operations. But here's your code

 int shift(int num, int digit) { return num*10 + digit; }

enjoy so

 int x = 1; x = shift(x, 2); cout << x;

You can of course rewrite and so

 void shift(int &num, int digit) { num num*10 + digit; }

and enjoy so

 int x = 1; shift(x, 2); cout << x;

but it is an amateur.

    You can do it like this, but I really wonder why?

     #include <iostream> using namespace std; class someclass{ private: int a; public: someclass():a(0) {} void operator =(int b){ a=a*10+b; } void show_a(){ cout<<a; } }; int main(){ someclass c; c=5; c=3; c.show_a(); return 0; }