c - allocating null pointer when passed to function -

this program contains typical node struct linked list (so pointer next node , int containing value). have following test functions:

void f(node** y, int value) {     node* x = *y;      if(!x)     {         printf("case1...\n");         x = (node*)malloc(sizeof(node));         x->data = value;         return;     }     printf("case2...\n");     x->next = (node*)malloc(sizeof(node));     x->next->data = value;     return; }  void myprint(node** y) {     node* x = *y;     printf("printing...\n");      printf("%d %d\n", x->data, x->next->data);     return; }  int main() {     node* n = null;      f(&n, 5);     f(&n, 10);      myprint(&n); } 

this code produces following output in linux:

case1... case1... printing... segmentation fault 

i don't understand why passing null pointer function cause first case occur. seems pointer being passed value, don't think that's what's going on. if call malloc() on node inside of main() , pass f(), second case hit never first. @ least makes partial sense me, since node never null when passed f() main(), allocating node before passing f() means null check inside f() never true.

is i'm trying possible? there way pass n f() while it's null , have behave way want to? or have allocate n outside of f() , remove null check inside of f()?

after allocation x need set in *y in function f(). otherwise, not reflected once f() returns.

so change code to

void f(node** y, int value) {     node* x = *y;      if(!x)     {         printf("case1...\n");         x = (node*)malloc(sizeof(node));         x->data = value;          *y = x;  //set allocated pointer         return;     }     printf("case2...\n");     x->next = (node*)malloc(sizeof(node));     x->next->data = value;     return; }