IBM PC case
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Suppose we need to create a segment occupying addresses F000:1000..F000:2000 Let's calculate linear addresses:
start = (0xF000 << 4) + 0x1000 = 0xF1000 end = (0xF000 << 4) + 0x2000 = 0xF2000The segment base must be selected so that the first offset in our segment will be 0x1000. Let's find it using the following equation:
VirtualAddress = LinearAddress - (SegmentBase << 4); 0x1000 = 0xF1000 - (base << 4);After solving this equation, we see that the segment base is equal to 0xF000. (you see, this is really a very simple case :) )
Now, we can create a segment entering:
segment start address: 0xF1000 segment end address: 0xF2000 segment base: 0xF000Please note that the end address never belongs to the segment in IDA.
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