Dynamic vector allocation

One of the reasons for the confusion is that the code uses variable names difficult to understand what they mean. With better names one can better follow what is happening. The code does more than it should and falls into some traps.

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

int main() {
    int indice = 0, indiceParcial = 0;
    int *vetor = malloc(sizeof(int) * 10);
    while (1) {
        int valor;
        scanf("%d", &valor);
        if (valor == 0) break;
        vetor[indice++] = valor;
        if (indiceParcial++ == 9) {
            int *vetorAuxiliar = malloc(sizeof(int) * (indice + 10));
            memcpy(vetorAuxiliar, vetor, indice * sizeof(int));
            free(vetor);
            vetor = vetorAuxiliar;
            indiceParcial = 0;
        }
    }
    printf("\n");
    for (int i = 0; i < indice; i++) printf("%d  ", vetor[i]);
}

See running on ideone . And no Coding Ground . Also I put GitHub for future reference .

• Only declare the variable when you need it
• You do not need to calloc() and should not cast of the result.
• I'm not worrying about the test to see if the allocation spoke because in exercise this will not fail
• This is a typical loop without end since the output is in the core of it does not have to use condition
• Once the value is entered you must determine whether to continue or exit the loop
• The first thing you should do is to add the value to the vector
• Next, it analyzes if the vector has filled up and must proceed with a reallocation and copying
• When you should recreate you create based on the current size plus 10, according to the statement
• And use memcpy() which is a ready and reliable function to copy data. You use sizeof in it for the same reason you use it in malloc() , these functions operate based on char , then copy byte per byte, and if you have a number of elements of a multibyte type you have to multiply by that number
• You can not forget to free the memory of the previous vector
• Then the new vector created as the default is copied by copying its address ( vetor becomes null after free() , if all goes well)
• And we zero the partial index that controls when it breaks the limit. It is possible to eliminate this variable and make a indice % 10 , but it will almost certainly be slower since the division is by far the slowest operation of the processor. Or use one thing or another, you need to decide whether you want simpler code or more performance.
• Maybe it would be better to use a for by reducing the scope of 2 variables.

Following your style ...

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

int main(){
    int *vet, c=0, c2=0, *vet1, aux, N=1;

    vet =(int*) calloc(10, sizeof(int));

    do {

        scanf("%d", &aux);
        if(!aux) break;

        if (c%10 == 0){
            N++;
            vet1 = (int*) malloc (N*10 * sizeof(int));
            for (int i=0; i<(10*(N-1)); i++) vet1[i] = vet[i];    
            free(vet);
            vet = (int*) malloc (N*10 * sizeof(int));
            for (int i=0; i<(10*(N-1)); i++) vet[i] = vet1[i];
            free(vet1);
        }
        c2++;
        vet[c++] = aux;
    } while (aux);

    printf("\n\n");
    for (int i=0; i<c2; i++) printf("%d  ", vet[i]);
    printf("\n");
    free(vet1);
    free(vet);

    return 0;
}