Printing vector memory garbage

Question

How do I make a program not print vector trash?

For example, I created 2 vectors at the beginning of my program with 5 elements each. If the user type numbers above 10 stores in v1 if it is larger store in v2. However, when printing, if the user has not entered any number that meets these conditions, the program prints memory junk.

It would be possible to print only the vectors that contain values?

Program for example only:

#include <stdio.h>
#include <stdlib.h>
#define n 5

int main()
{   
    int v1[n], v2[n], a, i;
    for(i=0;i<n;i++)
    {
      printf("Digite: ");
      scanf("%d",&a);
     if(a<=5) v1[i] = a;
      else v2[i] = a;
          }

    for(i=0;i<n;i++)
    { 
        printf("%d ", v1[i]);
    }
    for(i=0;i<n;i++)
    { 
        printf("%d ", v2[i]);
    }
 

system("pause");
return 0;
}

Answer 1

There’s a conceptual problem in storage. And I don’t even know if I can do it right because the statement is bad.

Each vector needs to store only the numbers that are pertinent to them. Then each vector position moves forward as an item is placed there, so you need to count how many items have already been placed in each one and this count is the position you should place. So it doesn’t leave holes in the vectors and then you’ll know how many there are in each to display only the amount that has been filled.

In a more organized and readable way:

#include <stdio.h>
#include <stdlib.h>
#define MAX 5

int main() {   
    int menores[MAX], maiores[MAX], cont_menores = 0, cont_maiores = 0;
    for (int i = 0; i < MAX; i++) {
        printf("Digite: ");
        int numero;
        scanf("%d", &numero);
        if (numero > 10) maiores[cont_maiores++] = numero;
        else menores[cont_menores++] = numero;
    }
    for (int i = 0; i < cont_menores; i++) printf("%d ", menores[i]);
    printf("\n");
    for (int i = 0; i < cont_maiores; i++) printf("%d ", maiores[i]);
}

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