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Hi,

I noticed that the Interrupt Vector Table page in the wiki lacks a piece of information vital to the reading and modification of the IVT: It does not mention that the memory addresses stored in the interrupt vectors are in reverse.

For example, modifying the interrupt vector of interrupt 0x21 to point to 0x7650:0x7B4F:

According to the article, once the interrupt vector is written, it should look like this:
A simple, but probably not the most efficient way to do this in x86 (NASM) assembly, would be:
But I ran into a problem when doing this, and I discovered that when I invoked my interrupt, the CPU would jump somewhere seemingly arbitrary. After careful debugging and research, I discovered that the memory addresses stored in the interrupt vectors were in reverse.

Therefore, once the interrupt vector is written, it should look like this:
A simple, but probably not the most efficient way to do this in x86 (NASM) assembly, would be:
I would be grateful if someone could please add this information to the wiki, as it would save people like me a few hours of (unnecessary) debugging. I didn't attempt to add this information to the wiki myself because I wasn't sure of any standards and if it was even worth adding to the wiki.

Thanks,

-U238

You're making a mistake yourself: is exactly equivalent to and thus does not make any difference. If you disagree, read up on byteorders.


Perhaps there is something else wrong?

U238 wrote:But I ran into a problem when doing this, and I discovered that when I invoked my interrupt, the CPU would jump somewhere seemingly arbitrary. After careful debugging and research, I discovered that the memory addresses stored in the interrupt vectors were in reverse.

Well, the x86 architecture is little-endian, but that is not the problem here. The problem is that you are not writing the IVT at all. The IVT is not located at 76500H but at 00000H.