Array size set by variable

Question

I checked and I see no mistakes.

void exibe (int * , int);

int main(){

  int n;
  int m[n], i;

  printf("Digite o numero de elementos do array : ");
  scanf("%d",&n);
 
  for (i = 0; i < n; i++){
    printf("Digite o %d numero : ",i + 1);
    scanf("%d",&m[i]);
   }

  exibe(m,n);

  system("pause");
  return 0;
}

void exibe (int *m, int n){
   int i;

   for (i = 0, i < n; i++)
    printf("Numero %d = %d \n",m[i]);
 }

Answer 1

There are typos like , in place of ;, missing put two data in a text template to print that expects two different data and the biggest problem is that is trying to create a array with a data that does not exist yet (even exists but is something caught almost randomly in memory), that usually happens with this wrong idea that people learn that they have to declare all variables before starting the code, which would prevent this code pattern.

#include <stdio.h>

void exibe (int *m, int n) {
    for (int i = 0; i < n; i++) printf("Numero %d = %d \n", i, m[i]);
}

int main() {
    int n;
    printf("Digite o numero de elementos do array : ");
    scanf("%d", &n);
    int m[n];
    for (int i = 0; i < n; i++) {
        printf("Digite o %d numero : ", i + 1);
        scanf("%d", &m[i]);
    }
    exibe(m, n);
}

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

This has there its risk but nothing critical for an exercise, if someone put a very large number, bigger than the capacity of the stack gives error, then the number should be validated before.