I'm trying to make a program that verifies if a number meets two requirements:

1. the number has 4 digits

2. not all the digits are the same

I've been made it, and it looks good, but the code fails in a specific case.

When the user enters a number with all of the digits equal, like 1111, the first time the program will say that this number isn't correct and will ask for a new number, but if the user enters another invalid number with all digits equal, like 2222, the verification fails and accepts this new number as a valid number.

I don't know what the problem could be, can someone help?

This is the code:

#include <iostream>
using namespace std;

int main(){
    
    int n; // number 
    cout << "enter a 4-digit number with at least one different digit number: ";
    cin >> n;
    
    int m = 0, o = 0, u = n, z = 0; // variables
    int counter, counter1, counter2 = 5; //counters 
    int x[4], y[4]; // vectors 
    
    //--------------------------- verification -------------------------------------
    while(counter != 4 && counter2 > 3){ 
        // ======= number size verification =======
        while (u > 0){
            u = u/10;
            counter = counter   1; // number digit counter
        }
        
        if(counter != 4){ // if the number has less than 4 digits
            cout << "the number has less than 4 digits, enter a valid number: ";
            cin >> n;
            
            u = n;
        }
        // =================================================
        
        // ========= check repetead numbers ================
        if(counter == 4){ // if the number has 4 digits
            u = n;
            z = n;
            for(int i = 3; i >= 0; i--){ // convert the number into two vectors
                u = u % 10;
                x[i] = u;
                y[i] = u;
                z = z / 10;
                u = z;
            }
            
            counter2 = 0; // repetition counter 2 (reset counter value)
            
            // traversal of the vector in search of repeated numbers
            for(int j = 0; j < 4; j  ){
              counter1 = 0; // repetition counter 1 (reset counter value) 
                m = x[j];
                for(int k = 0; k < 4; k  ){
                  o = y[k];
                    if(m == o){ 
                        counter1 = counter1   1; // if there are repeated numbers counter increases
                        y[k] = 0; // delete that vector position
                    }
                }
                
                if(counter1 == 1 || counter1 == 0){ // if only found the repetition of the same box
                  counter1 = 0; // reset counter1
                }else{
                  counter2 = counter2   counter1; // if not, its value is assigned to counter2
                }
            }
            

        }
        
        if(counter2 == 4){ // if the number has all its digits the same
            n = 0;
            cout << "the number has all its digits the same, enter a valid number: ";
            cin >> n;
        }
        
    }
    
    cout << "valid number";
    
    return 0;
}

image of the failure:

CodePudding user response：

To check if your number is 4 digits you can simply compare it:

if( n < 1000 or n > 9999 ) // not 4 digits number

To check if all digits are the same, just check if it is divisible by 1111 (thanks to @SamVarshavchik )

if( n % 1111 == 0 ) // all digits are the same