With the introduction of void * pointers in the ANSI/ISO C Standard, explicitly casting the result of a call to malloc is no longer necessary and may even produce unexpected behavior if #include <stdlib.h> is forgotten.
Non-Compliant Code Example
If stdlib.h is not included, the compiler makes the assumption that malloc has a return type of int. When the result of a call to malloc is then explicitly cast to a pointer type, the compiler will assume that the cast from int to a pointer type is done with full knowledge of the possible outcomes. This may lead to behavior which is unexpected by the programmer.
char *p = (char *)malloc(10);
Compliant Solution
By ommiting the explicit cast to a pointer the compiler will realise that an int is attempting to be assigned to a pointer type and will generate a warning which may easily be corrected.
#include <stdlib.h> ... char *p = malloc(10);
References
comp.lang.c FAQ list - Question 7.7 http://c-faq.com/malloc/cast.html
comp.lang.c FAQ list - Question 7.7b http://c-faq.com/malloc/mallocnocast.html