Programming Tips - How does libical allocate memory? Who owns it?
Date: 2010may17 Library: libical Q. How does libical allocate memory? Who owns it? A. The function name determines who (libical or the caller) owns the memory. Here is the relevant section excerpted from ~/doc/UsingLibical.txt in version 0.44 If that's not current, you can search for the first sentence: http://www.google.com/search?q=Libical+relies+heavily+on+dynamic+allocationMemory Management Libical relies heavily on dynamic allocation for both the core objects and for the strings used to hold values. Some of this memory the library caller owns and must free, and some of the memory is managed by the library. Here is a summary of the memory rules. 1) If the function name has "new" in it, the caller gets control of the memory. ( such as icalcomponent_new(), or icalproperty_new_clone() ) 2) If you got the memory from a routine with new in it, you must call the corresponding *_free routine to free the memory. ( Use icalcomponent_free() to free objects created with icalcomponent_new() ) 3) If the function name has "add" in it, the caller is transferring control of the memory to the routine. ( icalproperty_add_parameter() ) 4) If the function name has "remove" in it, the caller passes in a pointer to an object and after the call returns, the caller owns the object. So, before you call icalcomponent_remove_property(comp,foo), you do not own "foo" and after the call returns, you do. 5) If the routine returns a string and its name does NOT end in "_r", libical owns the memory and will put it on a ring buffer to reclaim later. For example, icalcomponent_as_ical_string(). You'd better strdup() it if you want to keep it, and you don't have to delete it. 6) If the routine returns a string and its name *does* end in "_r", the caller gets control of the memory and is responsible for freeing it. For example, icalcomponent_as_ical_string_r() does the same thing as icalcomponent_as_ical_string(), except you now have control of the string buffer it returns.