No. You don't need to.
Pointer arithmetic is performed based on the type of the pointer T *
. Adding a size_t
will not affect the pointer arithmetic as the increment is done using sizeof(T)
.
To quote the standard (C11 draft):
6.5.6 Additive operators
When an expression that has integer type is added to or subtracted
from a pointer, the result has the type of the pointer operand. If the
pointer operand points to an element of an array object, and the array
is large enough, the result points to an element offset from the
original element such that the difference of the subscripts of the
resulting and original array elements equals the integer expression.
In other words, if the expression P points to the i-th element of an
array object, the expressions (P)+N (equivalently, N+(P)) and (P)-N
(where N has the value n) point to, respectively, the i+n-th and
i−n-th elements of the array object, provided they exist. Moreover, if
the expression P points to the last element of an array object, the
expression (P)+1 points one past the last element of the array object,
and if the expression Q points one past the last element of an array
object, the expression (Q)-1 points to the last element of the array
object. If both the pointer operand and the result point to elements
of the same array object, or one past the last element of the array
object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined. If the result points one past the last element
of the array object, it shall not be used as the operand of a unary *
operator that is evaluated.
On the other hand, casting size_t
to ptrdiff_t
could lead incorrect code as ptrdiff_t
is a signed while size_t
is an unsigned type. So if the resulting value is larger than what ptrdiff_t
can hold then there's a problem. In short, pointer arithmetic is well-defined when adding any integral type to a pointer type and you don't need such a cast at all.