You have a misconception about the way C handles arguments. In C, all arguments are passed by value, including pointers. When you reassign a pointer inside of a function you are reassigning a copy of that pointer. For instance: void f ( int *p ); int *p; f(p); The address ( &p) of the pointer is different in the function. They both point to the same location (have the same value), but each has a different address. When you assign the pointer to the return value of malloc, it is only assigning the function local copy of that pointer. One way to fix this is to introduce another level of indirection, and pass the address of the pointer: void insertNode(int i, node **n), which you can call like insertNode(0, &n).
When you want to change it to something else, dereference it once and then assign: *p = malloc(sizeof(node)). Another solution is to have the function return the pointer and assign it in the calling code: return malloc(sizeof(node)). (Note: You would actually return it after the initialization code. Download Camtasia Studio 7 Full Crack Mf more. Also don't cast the return value of malloc in C).
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