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

I know that it's a common convention to pass the length of dynamically allocated arrays to functions that manipulate them:

void initializeAndFree(int* anArray, size_t length);

int main(){
    size_t arrayLength = 0;
    scanf("%d", &arrayLength);
    int* myArray = (int*)malloc(sizeof(int)*arrayLength);

    initializeAndFree(myArray, arrayLength);
}

void initializeAndFree(int* anArray, size_t length){
    int i = 0;
    for (i = 0; i < length; i++) {
        anArray[i] = 0;
    }
    free(anArray);
}

but if there's no way for me to get the length of the allocated memory from a pointer, how does free() "automagically" know what to deallocate when all I'm giving it is the very same pointer? Why can't I get in on the magic, as a C programmer?

Where does free() get its free (har-har) knowledge from?

Solution 1:

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.

Solution 2:

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.