I have a strcture called NODE to hold my list and a function NODE find() that looks for an item in the list and returns the NODE. How would I return an error code if the item isn't found? If I just put return -1 or return NULL I get an incompatible type error, and I don't know how to cast to the current type. (NODE *)NULL doesn't work.
you can only return types you can't return to a specific variable. exI guess you could return all elements of the structure to NULL or its equivalent.Code:#include <stdio.h> struct bob{ int x; }; struct bob func(int y,struct bob test); int main(void){ struct bob new; new=func(10,new); return 0; } struct bob func(int y, struct bob test){ test.x=y; return test; }
Last edited by linuxdude; 11-25-2004 at 10:59 PM.
You should be using pointers rather than passing structs around. It is far more efficient and you can return NULL when required. It also prevents problems if the original changes and is the accepted style.
Sorry, I didn't explain very well.
Code:typedef struct{ int bob, back, front; }NODE; // this is what I meant typedef struct{ NODE *list; //.... }LIST; // .............. NODE find(LIST l, int find_me){ //can't find "find_me" return /*NULL error code*/; // stuck here }
pass the pointer to a struct instead of the actual struct like anonytmous said
While you're at it, return a NODE pointer instead of a NODE.
Quzah.
Hope is the first step on the road to disappointment.
