void pointer as argument [duplicate]

Answer 1

As this is C, you cannot pass the pointer by reference without passing in a pointer to the pointer (e.g., void ** rather than void * to point to the pointer). You need to return the new pointer. What is happening:

f(a1);

Pushes the value of the pointer (NULL) as the stack parameter value for a. a picks up this value, and then reassigns itself a new value (the malloced address). As it was passed by value, nothing changes for a1.

If this were C++, you could achieve what you wanted by doing passing the pointer by reference:

void f(void *&a);