Convert for to while

Question

#include <stdio.h>
#include <stdlib.h>


int main(void) {

   int X[10], cont;
   cont=0;
   while(cont<10){
        printf("Digite um numero inteiro: ");
        scanf("%d",&X[cont]);
        if(X[cont] %2 == 0){
            X[cont]=0;
        }else{
            X[cont]=1;

        }
        cont++;
    }
    
   while(cont<10){
        printf(" %d ", X[cont]);
        cont++;
   }

    
}

This code is not working, but when I do for works:

#include <stdio.h>
#include <stdlib.h>


int main(void) {

   int X[10], cont;
   for(cont=0;cont<10;cont++){
        printf("Digite um numero inteiro: ");
        scanf("%d",&X[cont]);
        if(X[cont] %2 == 0){
            X[cont]=0;
        }else{
            X[cont]=1;

        }
    }
    
   for(cont=0;cont<10;cont++){
        printf(" %d ", X[cont]);
   }

    
}

Answer 1

The main problem is that it is not zeroing the counter when going to do the second loop, as the variable was ditching 10 at the output of the first it will pick wrong positions of memory in the sequence. I took the opportunity to organize and simplify the code.

Note that the result of the account to find out if it is even or odd already gives exactly the required number, so there is no reason to make a if.

#include <stdio.h>

int main(void) {
   int X[10];
   int cont = 0;
   while (cont < 10) {
        printf("Digite um numero inteiro: ");
        scanf("%d", &X[cont]);
        X[cont] = X[cont] % 2;
        cont++;
    }
    cont = 0;
    while (cont < 10) printf(" %d ", X[cont++]);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.