在堆上分配小于 4KB 的空间？

Question

Please Notice my followup question under accepted answer.

I just read about how translating from virtual to physical memory is done and I got an interesting question.

If the minimum size of frame in physical memory is 4KB why doesn't that say we can use heap for a minimum allocation of 4KB? Or that if I wanted only 1 byte on heap then it allocates the whole frame of 4KB and the rest is kept unused (even that there is space for others) Or It's split with other allocations/process since we have the offset field?

Answer 1

Try this code:

int main()
{
    int* a = malloc(sizeof *a);
    int* b = malloc(sizeof *b);
    printf("%p\n", (void*)a);
    printf("%p\n", (void*)b);
    return 0;
}

Possible output:

0x2164010
0x2164030

As you can see the allocated memory is within the same 4KB page (0x2164xxx), it has a 16-byte offset from the page and a 16-byte alignment).

Even if the mapping from physical to virtual addresses is done on some page size (eg 4KB), there may be several dynamic allocations within one mapped page.