I want to allocate a 2.9GB char array with
database = (char*) malloc((2900 * 1000000 * sizeof(char)));
This gives an integer overflow warning and the malloc
returns NULL
. The
malloc
parameter is of type size_t
which according to documentation is of type
unsigned int
.
So the max should be UINT_MAX
which is at least 2.9GB. However, if
I try to allocate more than MAX_INT
the malloc
fails. Does this mean
size_t
on my system is of type int? How do I check this? I looked through
/usr/include/stdlib.h
and
./lib/gcc/x86_64-redhat-linux/4.1.1/include/stddef.h
but
can't find the definition of size_t
. Thanks very much
There are two issues here.
First, the overflow warning: both 2900
and 1000000
are of type int
, so the result of multiplying them is also of type int
. The result cannot be represented by a 32-bit signed integer, so it overflows. You need to cast one (or both) arguments to size_t
to use unsigned arithmetic.
(Or, you could move the sizeof(char)
to be one of the first two terms, since its type is size_t
, though you can also just remove the sizeof(char)
since it is always 1
.)
Second, the maximum size that malloc
can allocate depends both on the platform on which you are running and on the current state of the program. If there is insufficient contiguous address space left to satisfy the request, obviously the malloc
will fail.
Further, the platform on which you are running may have an upper limit on how large an object it can dynamically allocate. You'll need to consult your platform's documentation to find out what that upper limit is.
size_t
is certainly not int
, because int
is always signed and size_t
is always unsigned.