From http://en.wikipedia.org/wiki/Malloc#Casting_and_type_safety
" Casting and type safety
malloc returns a void pointer (void *), which indicates that it is a
pointer to a region of unknown data type. One may "cast" (see type
conversion) this pointer to a specific type, as in
int *ptr = (int*)malloc(10 * sizeof (int));
When using C, this is considered bad practice; it is redundant under the
C standard. Moreover, putting in a cast may mask failure to include the
header stdlib.h, in which the prototype for malloc is found. In the
absence of a prototype for malloc, the C compiler will assume that
malloc returns an int, and will issue a warning in a context such as the
above, provided the error is not masked by a cast. On certain
architectures and data models (such as LP64 on 64 bit systems, where
long and pointers are 64 bit and int is 32 bit), this error can actually
result in undefined behavior, as the implicitly declared malloc returns
a 32 bit value whereas the actually defined function returns a 64 bit
value. Depending on calling conventions and memory layout, this may
result in stack smashing.
The returned pointer need not be explicitly cast to a more specific
pointer type, since ANSI C defines an implicit conversion between the
void pointer type and other pointers to objects. An explicit cast of
malloc's return value is sometimes performed because malloc originally
returned a char *, but this cast is unnecessary in standard C
code.[4][5] Omitting the cast, however, creates an incompatibility with
C++, which does require it.
The lack of a specific pointer type returned from malloc is type-unsafe
behaviour: malloc allocates based on byte count but not on type. This
distinguishes it from the C++ new operator that returns a pointer whose
type relies on the operand. (see C Type Safety). "
See also
http://www.opensc-project.org/pipermail/opensc-devel/2010-August/014586.html
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