Malloc inside a function call appears to be getting freed on return?

Question

I think I\'ve got it down to the most basic case:

int main(int argc, char ** argv) {
  int * arr;

  foo(arr);
  printf(\"car[3]=%d\\n\",arr[3]);
  free (arr         


        

Answer 1

You pass the pointer by value, not by reference, so whatever you do with arr inside foo will not make a difference outside the foo-function. As m_pGladiator wrote one way is to declare a reference to pointer like this (only possible in C++ btw. C does not know about references):

int main(int argc, char ** argv) {
  int * arr;

  foo(arr);
  printf("car[3]=%d\n",arr[3]);
  free (arr);
  return 1;
}

void foo(int * &arr ) {
  arr = (int*) malloc( sizeof(int)*25 );
  arr[3] = 69;
}

Another (better imho) way is to not pass the pointer as an argument but to return a pointer:

int main(int argc, char ** argv) {
  int * arr;

  arr = foo();
  printf("car[3]=%d\n",arr[3]);
  free (arr);
  return 1;
}

int * foo(void ) {
  int * arr;
  arr = (int*) malloc( sizeof(int)*25 );
  arr[3] = 69;
  return arr;
}

And you can pass a pointer to a pointer. That's the C way to pass by reference. Complicates the syntax a bit but well - that's how C is...

int main(int argc, char ** argv) {
  int * arr;

  foo(&arr);
  printf("car[3]=%d\n",arr[3]);
  free (arr);
  return 1;
}

void foo(int ** arr ) {
  (*arr) = (int*) malloc( sizeof(int)*25 );
  (*arr)[3] = 69;
}