understand passing parameters by reference with dynamic allocation

Question

I\'m trying understand how to pass a parameter by reference in C language. So I wrote this code to test the behavior of parameters passing:

#include 

        

Answer 1

C is pass-by-value, it doesn't provide pass-by-reference. In your case, the pointer (not what it points to) is copied to the function paramer (the pointer is passed by value - the value of a pointer is an address)

void alocar(int* n){
   //n is just a local variable here.
   n = (int*) malloc( sizeof(int));
  //assigning to n just assigns to the local
  //n variable, the caller is not affected.

You'd want something like:

int *alocar(void){
   int *n = malloc( sizeof(int));
   if( n == NULL )
      exit(-1);
   *n = 12;
   printf("%d.\n", *n);
   return n;
}
int main()
{
   int* n;
   n = alocar();
   printf("%d.\n", *n);
   return 0;
}

Or:

void alocar(int** n){
   *n =  malloc( sizeof(int));
   if( *n == NULL )
      exit(-1);
   **n = 12;
   printf("%d.\n", **n);
}
int main()
{
   int* n;
   alocar( &n );
   printf("%d.\n", *n);
   return 0;
}