Question

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

Answer 1

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.