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Hello.

I read http://wiki.osdev.org/Memory_Map_(x86)#Overview where you can see that from 0x00000000 to 0x000003FF we have IVT.

But I have written a memory detection code (INT 15h, EAX=e820) that I execute in the Bochs VM and I get these entries:



How is it possible? I mean, I have free memory from 0x0 to 0x9f000 but there we have the IVT, isn't it?

Thank you.

IVT (interrupt vector table - table with addresses of interrupt handlers in real mode) is only needed when you want interrupts in so-called real mode. When you leave this legacy mode, you can use this memory area for any general purpose. There are some exceptions (when you are going to use BIOS calls instead of some native drivers or when SMM handler is badly written and corrupts low memory) but most of the time this is just usual a memory region with nothing special about it.

The IVT isn't the only thing there. Everything below the first 1MB of physical address space, particularly the first 64K, is encumbered with legacy memory areas that the BIOS memory detection call assumes you know about due to the fact that every PC since the IBM 5150 has had the same memory scheme in the area from 0x0000 to 0xFFFF, which includes the BIOS ROM, peripheral option ROMs, legacy video memory, BIOS shadow memory and working memory, and a million other things.

Long story short: That table that you linked to (http://wiki.osdev.org/Memory_Map_(x86)#Overview) is always existent in every PC and non-negotiable. The BIOS memory detection interrupt is just there to tell you how all of the *other* memory is organized, because there's no standard for that like there is for the low memory area.

Ok. My doubt is solved because I was thinking I had done something wrong with memory detection . Thanks for comment.

The IVT memory area is probably not marked in the memory map as reserved because it is perfectly legal, and in most cases necessary for you to write to that area.

All of the other memory map area types are off limits to the OS, in some fashion.