Assembly 8086 addressing mode

Question

How does XOR [130][BX][DI], CL work? (I know what XOR does) I mean how is the effective address of the memory part calculated? Which addressing mode is this?

Answer 1

You wrote [130] , not 130 . Are you sure that's correct? I don't know whether (and I don't think) that's possible.

"The Art of Assembly Language" calls this "Based Indexed Plus Displacement Addressing Mode" .

The default segment in this case is DS (because you use BX ), which means your instruction is the same as

XOR DS:130[BX][DI], CL

The effective address is calculated by taking the segment address and adding 130 , BX , and DI . So it would be (DS*10h)+130+BX+DI .