Needless pointer-casts in C
c, c++, pointers
Solution
It seems fitting I post an answer, since I left the comment :P
Basically, if you forget to include `stdlib.h` the compiler will assume `malloc` returns an `int`. Without casting, you will get a warning. With casting you won't.
So by casting you get nothing, and run the risk of suppressing legitimate warnings.
Much is written about this, a quick google search will turn up more detailed explanations.
edit
It has been argued that
TYPE * p;
p = (TYPE *)malloc(n*sizeof(TYPE));
makes it obvious when you accidentally don't allocate enough memory because say, you thought `p` was `TYPe` not `TYPE`, and thus we should cast malloc because the advantage of this method overrides the smaller cost of accidentally suppressing compiler warnings.
I would like to point out 2 things:
- you should write `p = malloc(sizeof(*p)*n);` to always ensure you malloc the right amount of space
- with the above approach, you need to make changes in 3 places if you ever change the type of `p`: once in the declaration, once in the `malloc`, and once in the cast.
In short, I still personally believe there is no need for casting the return value of `malloc` and it is certainly not best practice.
Problem
I got a comment to my answer on this thread: Malloc inside a function call appears to be getting freed on return? In short I had code like this: ``` int * somefunc (void) { int * temp = (int*) malloc (sizeof (int)); temp[0] = 0; return temp; } ``` I got this comment: Can I just say, please don't cast the return value of malloc? It is not required and can hide errors. I agree that the cast is not required in C. It is mandatory in C++, so I usually add them just in case I have to port the code in C++ one day. However, I wonder how casts like this can hide errors. Any ideas? Edit: Seems like there are very good and valid arguments on both sides. Thanks for posting, folks.