I am making a function that will return a pointer to a long long variable. For example, I have the next variable prototype:
I want to change the return value, and the first parameter to pointers to long long.Code:long long funcName(long long x, int s);
But if you don't even know the most basic pointer syntax, you are almost certainly going to misuse the pointers themselves, and cause some sort of difficult-to-debug problem that may result in program crashes and strange behavior. I strongly recommend reading up on pointers in your C textbook, or on some good online tutorials (like the tutorial on this site, for example).Code:long long *funcName(long long *x, int s)
By "change the return value", you may not need the function to return a pointer nor to accept a pointer. The caller can do that.
Code:long long retval; retval = funcName(2LL, 3); /* call consistent with the OP's original code */ retval += 42; /* we can change retval here */
You only need the function to work with pointers if the function needs to change variables passed by the caller.
So, if I use it on malloc(), would it be:
?Code:malloc(*(long long));
But that gives me the error, "error: expected expression before ')' token"
No, it would be:
You allocate space for what the pointer points to, not the pointer itself (which already exists anyway).Code:x = malloc(sizeof(*x));
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
