If free() knows the length of my array, why can't I ask for it in my own code?

Besides Klatchko's correct point that the standard does not provide for it, real malloc/free implementations often allocate more space then you ask for. E.g. if you ask for 12 bytes it may provide 16 (see A Memory Allocator, which notes that 16 is a common size). So it doesn't need to know you asked for 12 bytes, just that it gave you a 16-byte chunk.

You can't get it because the C committee did not require that in the standard.

If you are willing to write some non-portable code, you may have luck with:

*((size_t *)ptr - 1)

or maybe:

*((size_t *)ptr - 2)

But whether that works will depend on exactly where the implementation of malloc you are using stores that data.

While it is possible to get the meta-data that the memory allocator places preceding the allocated block, this would only work if the pointer is truly a pointer to a dynamically allocated block. This would seriously affect the utility of function requiring that all passed arguments were pointers to such blocks rather than say a simple auto or static array.

The point is there is no portable way from inspection of the pointer to know what type of memory it points to. So while it is an interesting idea, it is not a particularly safe proposition.

A method that is safe and portable would be to reserve the first word of the allocation to hold the length. GCC (and perhaps some other compilers) supports a non-portable method of implementing this using a structure with a zero length array which simplifies the code somewhat compared to a portable solution:

typedef tSizedAlloc
{
    size_t length ;
    char* alloc[0] ;   // Compiler specific extension!!!
} ;

// Allocating a sized block
tSizedAlloc* blk = malloc( sizeof(tSizedAlloc) + length ) ;
blk->length = length ;

// Accessing the size and data information of the block
size_t blk_length = blk->length ;
char*  data = blk->alloc ;