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);   return 1; }  void foo(int * arr) {   arr = (int*) malloc( sizeof(int)*25 );   arr[3] = 69; } 

The output is this:

> ./a.out   car[3]=-1869558540  a.out(4100) malloc: *** error for object 0x8fe01037: Non-aligned pointer                          being freed  *** set a breakpoint in malloc_error_break to debug > 

If anyone can shed light on where my understanding is failing, it’d be greatly appreciated.

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; }