When a pointer to a particular type (say int
, char
, float
, ..) is incremented, its value is increased by the size of that data type. If a void
pointer which points to data of size x
is incremented, how does it get to point x
bytes ahead? How does the compiler know to add x
to value of the pointer?
Final conclusion: arithmetic on a void*
is illegal in both C and C++.
GCC allows it as an extension, see Arithmetic on void
- and Function-Pointers (note that this section is part of the "C Extensions" chapter of the manual). Clang and ICC likely allow void*
arithmetic for the purposes of compatibility with GCC. Other compilers (such as MSVC) disallow arithmetic on void*
, and GCC disallows it if the -pedantic-errors
flag is specified, or if the -Werror-pointer-arith
flag is specified (this flag is useful if your code base must also compile with MSVC).
Quotes are taken from the n1256 draft.
The standard's description of the addition operation states:
6.5.6-2: For addition, either both operands shall have arithmetic type, or one operand shall be a pointer to an object type and the other shall have integer type.
So, the question here is whether void*
is a pointer to an "object type", or equivalently, whether void
is an "object type". The definition for "object type" is:
6.2.5.1: Types are partitioned into object types (types that fully describe objects) , function types (types that describe functions), and incomplete types (types that describe objects but lack information needed to determine their sizes).
And the standard defines void
as:
6.2.5-19: The
void
type comprises an empty set of values; it is an incomplete type that cannot be completed.
Since void
is an incomplete type, it is not an object type. Therefore it is not a valid operand to an addition operation.
Therefore you cannot perform pointer arithmetic on a void
pointer.
Originally, it was thought that void*
arithmetic was permitted, because of these sections of the C standard:
6.2.5-27: A pointer to void shall have the same representation and alignment requirements as a pointer to a character type.
However,
The same representation and alignment requirements are meant to imply interchangeability as arguments to functions, return values from functions, and members of unions.
So this means that printf("%s", x)
has the same meaning whether x
has type char*
or void*
, but it does not mean that you can do arithmetic on a void*
.
Editor's note: This answer has been edited to reflect the final conclusion.