这个返回值的目的是什么？

Question

I ran into some code I couldn't find an answer to on Google or SO. I am looking at a thread function which returns void* as you could expect. However, before the thread function ends it suddenly pulls this stunt,

return (void*) 0;

What is the purpose of that? I can't make any sense of it.

edit:

After understanding this is the same as NULL-- it is my thought they used this to skip including stdlib.

Answer 1

(void*)0 is the null pointer, aka NULL (which actually is a macro defined in several header files, eg stddef.h or stdio.h , that basically amounts to the same thing as (void*)0 ).

Update:

How to explain null pointers and their usefulness? Basically, it's a special value that says, "This pointer doesn't point anywhere," or, "This pointer is not set to a valid object reference."

Historical note: Tony Hoare, who is said to have invented null references in 1965, is known to regret that invention and thus calls it his "Billion Dollar Mistake" :

Whenever you work with pointers, you must make sure to never dereference a null pointer (because it doesn't reference anything by definition). If you do it anyway, you'll either get abnormal program termination, a general protection fault, or unexpected program behaviour at the very least.

Answer 2

Well, I have not encountered any C++ compiler saying NULL or 0 cannot be converted to void* (or to/from int* , for example). But there might be some smart compilers or static-analysis tools that would report 0 to void-pointer conversion as a warning.

That statement is commonly found in callback implementation (like a thread-routine), which must adhere to prototype of callback being demanded ( pthread_create , CreateThread etc). Therefore, when you implement that function, you must return the same type it was demanded for. For pthread_create routine , you must return a void* - and that's why return (void*)0; is there.