c++ - "if" statement with "or" operator not working -

trying have program make user renter input if choose number isn't 1 2 or 3.
whenever type 1 2 or 3 still wants me re enter integer. did use or operator wrong?

#include <iostream> #include <string>   using namespace std;   int main() {      cout << "choose option 1,2 or 3: \n";     cout << "1: drop single chip 1 slot \n";     cout << "2: drop multiple chips 1 slot \n";     cout << "3: quit program \n";      int choice;     cin >> choice;       if  (choice != 1||2||3)     {         cout << "please enter 1 2 or 3 \n";         cin >> choice;     }          else               {                 cout << "it worked \n";             } } 

this condition in if statement

if  (choice != 1||2||3) 

is equivalent to

if  ( ( choice != 1 ) || 2 || 3 ) 

so equal true because if choice != 1 evaluates false nevertheless expressions 2 , 3 unequal 0 , consequently each of them evaluate true.

what mean following

if  ( ( choice != 1 ) && ( choice != 2 ) && ( choice != 3 ) ) 

or write simpler

if  ( choice < 1  || choice > 3 ) 

or maybe 1 more readable

if  ( !( 1 <= choice && choice <= 3 ) ) 

or following way:)

if  ( not ( 1 <= choice && choice <= 3 ) ) 

take account part of program enclose in do-while loop. example

int choice; bool valid_input;  {     cout << "choose option 1,2 or 3: \n";     cout << "1: drop single chip 1 slot \n";     cout << "2: drop multiple chips 1 slot \n";     cout << "3: quit program \n";      if ( !( cin >> choice ) ) break;      valid-input  = 1 <= choice && choice <= 3;      if  ( !valid_input )     {         cout << "please enter 1 2 or 3 \n";     } } while ( !valid_input );  if ( valid_input ) {     cout << "it worked. there selected option " << choice << std::endl; }