when explicit pointer type cast is needed?

Question

In C, if I have two types of pointers：

typeB *q;
typeA *p; 

int the following, is it unnecessary to put an explicit cast (typeA *)

1    p= q       is fine, no need to use p = (typeA  *)q;

2    define: void func(typeA *p){}

     call:  func(q);  is fine, no need to use func((typeA  *)q)

3    p=malloc(sizeof(typeA));   is fine, no need to use p=(typeA  *)malloc(sizeof(typeA))

besides, is C++ the same or not?

thanks!

Answer 1

I'll assume that they are unrelated types. If they are both aliases for the same type, then pointers to one can be implicitly converted to pointers to the other.

1. No; in both languages, you need a cast to convert between unrelated pointer types. In C++, it is fine if typeA is a base class of typeB.

2. Exactly the same as the first - both are trying to initialise a typeA* from a typeB*.

3. Fine in C; void* can be implicitly converted to any object pointer type. That conversion is not allowed in C++; but you usually shouldn't be using malloc anyway.

Answer 2

If you use TypeB * where TypeA * is expected, then you always need the cast. In C, there are two exceptions: conversion from and to void * is implicit (no cast needed) if the other type is a data pointer type. It's only under POSIX that void * is also implicitly compatible with function pointer types.

In C++, however, only conversion to void * is implicit, so assigning T *p = <expression of type void *>; still needs the cast.

Another difference is that in C++, if TypeA and TypeB are class (or struct) types, and TypeB inherits from TypeA then conversion from TypeB * to TypeA * ("downcasting") is again implicit (no cast needed). Of course this doesn't work backwards ("upcasting").

All in all, you don't cast the return value of malloc(), because:

• in C is not required;

• In C++ it would be required but in C++ you don't use malloc() anyway.